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Given the equation $9 x^2+9 y^2-4 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2-4 y+\left(9 x^2-10\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
9 y^2-4 y+9 x^2=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(9 y^2-4 y+\underline{\text{ }}\right)+9 x^2=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 y^2-4 y+\underline{\text{ }}\right)=9 \left(y^2-\frac{4 y}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(y^2-\frac{4 y}{9}+\underline{\text{ }}\right)$}+9 x^2=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{9}}{2}\right)^2=\frac{4}{81} \text{on }\text{the }\text{left }\text{and }9\times \frac{4}{81}=\frac{4}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
10+\frac{4}{9}=\frac{94}{9}: \\
9 \left(y^2-\frac{4 y}{9}+\frac{4}{81}\right)+9 x^2=\fbox{$\frac{94}{9}$} \\
\end{array}
Step 7:
\begin{array}{l}
y^2-\frac{4 y}{9}+\frac{4}{81}=\left(y-\frac{2}{9}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \fbox{$\left(y-\frac{2}{9}\right)^2$}+9 x^2=\frac{94}{9} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2-2 y^2+y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+y+\left(8 x^2+4\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-2 y^2+y+8 x^2+4 \text{from }\text{both }\text{sides}: \\
2 y^2-y+\left(-8 x^2-4\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
2 y^2-y-8 x^2=4 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(2 y^2-y+\underline{\text{ }}\right)-8 x^2=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2-y+\underline{\text{ }}\right)=2 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}-8 x^2=\underline{\text{ }}+4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{2}{16}=\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
4+\frac{1}{8}=\frac{33}{8}: \\
2 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)-8 x^2=\fbox{$\frac{33}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
y^2-\frac{y}{2}+\frac{1}{16}=\left(y-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$\left(y-\frac{1}{4}\right)^2$}-8 x^2=\frac{33}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2+x-3 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 x^2+x+(-3 y-8)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-3 y-5 x^2+x-8 \text{from }\text{both }\text{sides}: \\
5 x^2-x+(3 y+8)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }3 y+8 \text{from }\text{both }\text{sides}: \\
5 x^2-x=-3 y-8 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(5 x^2-x+\underline{\text{ }}\right)=(-3 y-8)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 x^2-x+\underline{\text{ }}\right)=5 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right)$}=(-3 y-8)+\underline{\text{ }} \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{5}{100}=\frac{1}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
(-3 y-8)+\frac{1}{20}=-3 y-\frac{159}{20}: \\
5 \left(x^2-\frac{x}{5}+\frac{1}{100}\right)=\fbox{$-3 y-\frac{159}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{x}{5}+\frac{1}{100}=\left(x-\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \fbox{$\left(x-\frac{1}{10}\right)^2$}=-3 y-\frac{159}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2+6 x-y^2+y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2+y+3 x^2+6 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-y^2+y+3 x^2+6 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2+6 x+\underline{\text{ }}\right)+\left(-y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2+6 x+\underline{\text{ }}\right)=3 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2+2 x+\underline{\text{ }}\right)$}+\left(-y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2+y+\underline{\text{ }}\right)=-\left(y^2-y+\underline{\text{ }}\right): \\
3 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$-\left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }3\times 1=3 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+3=9: \\
3 \left(x^2+2 x+1\right)-\left(y^2-y+\underline{\text{ }}\right)=\fbox{$9$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-1}{4}=-\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
9-\frac{1}{4}=\frac{35}{4}: \\
3 \left(x^2+2 x+1\right)-\left(y^2-y+\frac{1}{4}\right)=\fbox{$\frac{35}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
3 \fbox{$(x+1)^2$}-\left(y^2-y+\frac{1}{4}\right)=\frac{35}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 (x+1)^2-\fbox{$\left(y-\frac{1}{2}\right)^2$}=\frac{35}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2-5 x-4 y^2-3 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-3 y-4 x^2-5 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-4 y^2-3 y-4 x^2-5 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-4 x^2-5 x+\underline{\text{ }}\right)+\left(-4 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2-5 x+\underline{\text{ }}\right)=-4 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right)$}+\left(-4 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2-3 y+\underline{\text{ }}\right)=-4 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right): \\
-4 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }-4\times \frac{25}{64}=-\frac{25}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-1-\frac{25}{16}=-\frac{41}{16}: \\
-4 \left(x^2+\frac{5 x}{4}+\frac{25}{64}\right)-4 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right)=\fbox{$-\frac{41}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }-4\times \frac{9}{64}=-\frac{9}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{41}{16}-\frac{9}{16}=-\frac{25}{8}: \\
-4 \left(x^2+\frac{5 x}{4}+\frac{25}{64}\right)-4 \left(y^2+\frac{3 y}{4}+\frac{9}{64}\right)=\fbox{$-\frac{25}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{4}+\frac{25}{64}=\left(x+\frac{5}{8}\right)^2: \\
-4 \fbox{$\left(x+\frac{5}{8}\right)^2$}-4 \left(y^2+\frac{3 y}{4}+\frac{9}{64}\right)=-\frac{25}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{4}+\frac{9}{64}=\left(y+\frac{3}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x+\frac{5}{8}\right)^2-4 \fbox{$\left(y+\frac{3}{8}\right)^2$}=-\frac{25}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2+8 x-2 y^2-y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-y-x^2+8 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-2 y^2-y-x^2+8 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2+8 x+\underline{\text{ }}\right)+\left(-2 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2+8 x+\underline{\text{ }}\right)=-\left(x^2-8 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2-8 x+\underline{\text{ }}\right)$}+\left(-2 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2-y+\underline{\text{ }}\right)=-2 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right): \\
-\left(x^2-8 x+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-8}{2}\right)^2=16 \text{on }\text{the }\text{left }\text{and }-16=-16 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9-16=-7: \\
-\left(x^2-8 x+16\right)-2 \left(y^2+\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$-7$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-2}{16}=-\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-7-\frac{1}{8}=-\frac{57}{8}: \\
-\left(x^2-8 x+16\right)-2 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=\fbox{$-\frac{57}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-8 x+16=(x-4)^2: \\
-\fbox{$(x-4)^2$}-2 \left(y^2+\frac{y}{2}+\frac{1}{16}\right)=-\frac{57}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{2}+\frac{1}{16}=\left(y+\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -(x-4)^2-2 \fbox{$\left(y+\frac{1}{4}\right)^2$}=-\frac{57}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2+10 x+7 y^2-2 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2-2 y+5 x^2+10 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
7 y^2-2 y+5 x^2+10 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2+10 x+\underline{\text{ }}\right)+\left(7 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+10 x+\underline{\text{ }}\right)=5 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+2 x+\underline{\text{ }}\right)$}+\left(7 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2-2 y+\underline{\text{ }}\right)=7 \left(y^2-\frac{2 y}{7}+\underline{\text{ }}\right): \\
5 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$7 \left(y^2-\frac{2 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }5\times 1=5 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+5=11: \\
5 \left(x^2+2 x+1\right)+7 \left(y^2-\frac{2 y}{7}+\underline{\text{ }}\right)=\fbox{$11$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{7}}{2}\right)^2=\frac{1}{49} \text{on }\text{the }\text{left }\text{and }\frac{7}{49}=\frac{1}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
11+\frac{1}{7}=\frac{78}{7}: \\
5 \left(x^2+2 x+1\right)+7 \left(y^2-\frac{2 y}{7}+\frac{1}{49}\right)=\fbox{$\frac{78}{7}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
5 \fbox{$(x+1)^2$}+7 \left(y^2-\frac{2 y}{7}+\frac{1}{49}\right)=\frac{78}{7} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{2 y}{7}+\frac{1}{49}=\left(y-\frac{1}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 (x+1)^2+7 \fbox{$\left(y-\frac{1}{7}\right)^2$}=\frac{78}{7} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+10 x-9 y^2-9 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-9 y+8 x^2+10 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-9 y^2-9 y+8 x^2+10 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(8 x^2+10 x+\underline{\text{ }}\right)+\left(-9 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2+10 x+\underline{\text{ }}\right)=8 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right)$}+\left(-9 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2-9 y+\underline{\text{ }}\right)=-9 \left(y^2+y+\underline{\text{ }}\right): \\
8 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+y+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }8\times \frac{25}{64}=\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{25}{8}-5=-\frac{15}{8}: \\
8 \left(x^2+\frac{5 x}{4}+\frac{25}{64}\right)-9 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$-\frac{15}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-9}{4}=-\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{15}{8}-\frac{9}{4}=-\frac{33}{8}: \\
8 \left(x^2+\frac{5 x}{4}+\frac{25}{64}\right)-9 \left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{33}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{4}+\frac{25}{64}=\left(x+\frac{5}{8}\right)^2: \\
8 \fbox{$\left(x+\frac{5}{8}\right)^2$}-9 \left(y^2+y+\frac{1}{4}\right)=-\frac{33}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x+\frac{5}{8}\right)^2-9 \fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{33}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2+6 x-9 y^2+5 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2+5 y+2 x^2+6 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-9 y^2+5 y+2 x^2+6 x=-8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2+6 x+\underline{\text{ }}\right)+\left(-9 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2+6 x+\underline{\text{ }}\right)=2 \left(x^2+3 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2+3 x+\underline{\text{ }}\right)$}+\left(-9 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2+5 y+\underline{\text{ }}\right)=-9 \left(y^2-\frac{5 y}{9}+\underline{\text{ }}\right): \\
2 \left(x^2+3 x+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2-\frac{5 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}-8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }2\times \frac{9}{4}=\frac{9}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{2}-8=-\frac{7}{2}: \\
2 \left(x^2+3 x+\frac{9}{4}\right)-9 \left(y^2-\frac{5 y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{7}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{9}}{2}\right)^2=\frac{25}{324} \text{on }\text{the }\text{left }\text{and }-9\times \frac{25}{324}=-\frac{25}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{7}{2}-\frac{25}{36}=-\frac{151}{36}: \\
2 \left(x^2+3 x+\frac{9}{4}\right)-9 \left(y^2-\frac{5 y}{9}+\frac{25}{324}\right)=\fbox{$-\frac{151}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+3 x+\frac{9}{4}=\left(x+\frac{3}{2}\right)^2: \\
2 \fbox{$\left(x+\frac{3}{2}\right)^2$}-9 \left(y^2-\frac{5 y}{9}+\frac{25}{324}\right)=-\frac{151}{36} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{5 y}{9}+\frac{25}{324}=\left(y-\frac{5}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x+\frac{3}{2}\right)^2-9 \fbox{$\left(y-\frac{5}{18}\right)^2$}=-\frac{151}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2+9 x+2 y^2-10 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-10 y+x^2+9 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
2 y^2-10 y+x^2+9 x=4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(x^2+9 x+\underline{\text{ }}\right)+\left(2 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 y^2-10 y+\underline{\text{ }}\right)=2 \left(y^2-5 y+\underline{\text{ }}\right): \\
\left(x^2+9 x+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-5 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{9}{2}\right)^2=\frac{81}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
4+\frac{81}{4}=\frac{97}{4}: \\
\left(x^2+9 x+\frac{81}{4}\right)+2 \left(y^2-5 y+\underline{\text{ }}\right)=\fbox{$\frac{97}{4}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }2\times \frac{25}{4}=\frac{25}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{97}{4}+\frac{25}{2}=\frac{147}{4}: \\
\left(x^2+9 x+\frac{81}{4}\right)+2 \left(y^2-5 y+\frac{25}{4}\right)=\fbox{$\frac{147}{4}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+9 x+\frac{81}{4}=\left(x+\frac{9}{2}\right)^2: \\
\fbox{$\left(x+\frac{9}{2}\right)^2$}+2 \left(y^2-5 y+\frac{25}{4}\right)=\frac{147}{4} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-5 y+\frac{25}{4}=\left(y-\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \left(x+\frac{9}{2}\right)^2+2 \fbox{$\left(y-\frac{5}{2}\right)^2$}=\frac{147}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2-9 x-2 y^2-10 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-10 y-9 x^2-9 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
-2 y^2-10 y-9 x^2-9 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-9 x^2-9 x+\underline{\text{ }}\right)+\left(-2 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-9 x^2-9 x+\underline{\text{ }}\right)=-9 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$-9 \left(x^2+x+\underline{\text{ }}\right)$}+\left(-2 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2-10 y+\underline{\text{ }}\right)=-2 \left(y^2+5 y+\underline{\text{ }}\right): \\
-9 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2+5 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-9}{4}=-\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5-\frac{9}{4}=\frac{11}{4}: \\
-9 \left(x^2+x+\frac{1}{4}\right)-2 \left(y^2+5 y+\underline{\text{ }}\right)=\fbox{$\frac{11}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }-2\times \frac{25}{4}=-\frac{25}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{11}{4}-\frac{25}{2}=-\frac{39}{4}: \\
-9 \left(x^2+x+\frac{1}{4}\right)-2 \left(y^2+5 y+\frac{25}{4}\right)=\fbox{$-\frac{39}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-9 \fbox{$\left(x+\frac{1}{2}\right)^2$}-2 \left(y^2+5 y+\frac{25}{4}\right)=-\frac{39}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+5 y+\frac{25}{4}=\left(y+\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -9 \left(x+\frac{1}{2}\right)^2-2 \fbox{$\left(y+\frac{5}{2}\right)^2$}=-\frac{39}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2+7 x+y^2-5 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2-5 y+3 x^2+7 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
y^2-5 y+3 x^2+7 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2+7 x+\underline{\text{ }}\right)+\left(y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2+7 x+\underline{\text{ }}\right)=3 \left(x^2+\frac{7 x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2+\frac{7 x}{3}+\underline{\text{ }}\right)$}+\left(y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{3}}{2}\right)^2=\frac{49}{36} \text{on }\text{the }\text{left }\text{and }3\times \frac{49}{36}=\frac{49}{12} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
9+\frac{49}{12}=\frac{157}{12}: \\
3 \left(x^2+\frac{7 x}{3}+\frac{49}{36}\right)+\left(y^2-5 y+\underline{\text{ }}\right)=\fbox{$\frac{157}{12}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-5}{2}\right)^2=\frac{25}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{157}{12}+\frac{25}{4}=\frac{58}{3}: \\
3 \left(x^2+\frac{7 x}{3}+\frac{49}{36}\right)+\left(y^2-5 y+\frac{25}{4}\right)=\fbox{$\frac{58}{3}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+\frac{7 x}{3}+\frac{49}{36}=\left(x+\frac{7}{6}\right)^2: \\
3 \fbox{$\left(x+\frac{7}{6}\right)^2$}+\left(y^2-5 y+\frac{25}{4}\right)=\frac{58}{3} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-5 y+\frac{25}{4}=\left(y-\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x+\frac{7}{6}\right)^2+\fbox{$\left(y-\frac{5}{2}\right)^2$}=\frac{58}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+2 x+4 y^2+6 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2+6 y+4 x^2+2 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
4 y^2+6 y+4 x^2+2 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2+2 x+\underline{\text{ }}\right)+\left(4 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2+2 x+\underline{\text{ }}\right)=4 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)$}+\left(4 y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2+6 y+\underline{\text{ }}\right)=4 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right): \\
4 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{4}{16}=\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10+\frac{1}{4}=\frac{41}{4}: \\
4 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+4 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{41}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }4\times \frac{9}{16}=\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{41}{4}+\frac{9}{4}=\frac{25}{2}: \\
4 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+4 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$\frac{25}{2}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{2}+\frac{1}{16}=\left(x+\frac{1}{4}\right)^2: \\
4 \fbox{$\left(x+\frac{1}{4}\right)^2$}+4 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\frac{25}{2} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{2}+\frac{9}{16}=\left(y+\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x+\frac{1}{4}\right)^2+4 \fbox{$\left(y+\frac{3}{4}\right)^2$}=\frac{25}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2+7 x+y^2-9 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2-9 y-10 x^2+7 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
y^2-9 y-10 x^2+7 x=-8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2+7 x+\underline{\text{ }}\right)+\left(y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2+7 x+\underline{\text{ }}\right)=-10 \left(x^2-\frac{7 x}{10}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2-\frac{7 x}{10}+\underline{\text{ }}\right)$}+\left(y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{10}}{2}\right)^2=\frac{49}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{49}{400}=-\frac{49}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
-8-\frac{49}{40}=-\frac{369}{40}: \\
-10 \left(x^2-\frac{7 x}{10}+\frac{49}{400}\right)+\left(y^2-9 y+\underline{\text{ }}\right)=\fbox{$-\frac{369}{40}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-9}{2}\right)^2=\frac{81}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{81}{4}-\frac{369}{40}=\frac{441}{40}: \\
-10 \left(x^2-\frac{7 x}{10}+\frac{49}{400}\right)+\left(y^2-9 y+\frac{81}{4}\right)=\fbox{$\frac{441}{40}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-\frac{7 x}{10}+\frac{49}{400}=\left(x-\frac{7}{20}\right)^2: \\
-10 \fbox{$\left(x-\frac{7}{20}\right)^2$}+\left(y^2-9 y+\frac{81}{4}\right)=\frac{441}{40} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-9 y+\frac{81}{4}=\left(y-\frac{9}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x-\frac{7}{20}\right)^2+\fbox{$\left(y-\frac{9}{2}\right)^2$}=\frac{441}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2+3 x+6 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 x^2+3 x+(6 y-2)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 y-10 x^2+3 x-2 \text{from }\text{both }\text{sides}: \\
10 x^2-3 x+(2-6 y)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }2-6 y \text{from }\text{both }\text{sides}: \\
10 x^2-3 x=6 y-2 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(10 x^2-3 x+\underline{\text{ }}\right)=(6 y-2)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 x^2-3 x+\underline{\text{ }}\right)=10 \left(x^2-\frac{3 x}{10}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2-\frac{3 x}{10}+\underline{\text{ }}\right)$}=(6 y-2)+\underline{\text{ }} \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{10}}{2}\right)^2=\frac{9}{400} \text{on }\text{the }\text{left }\text{and }10\times \frac{9}{400}=\frac{9}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
(6 y-2)+\frac{9}{40}=6 y-\frac{71}{40}: \\
10 \left(x^2-\frac{3 x}{10}+\frac{9}{400}\right)=\fbox{$6 y-\frac{71}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{3 x}{10}+\frac{9}{400}=\left(x-\frac{3}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \text{10 }\fbox{$\left(x-\frac{3}{20}\right)^2$}=6 y-\frac{71}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2+9 x+7 y^2+5 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
7 y^2+5 y+9 x^2+9 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
7 y^2+5 y+9 x^2+9 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(9 x^2+9 x+\underline{\text{ }}\right)+\left(7 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2+9 x+\underline{\text{ }}\right)=9 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2+x+\underline{\text{ }}\right)$}+\left(7 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(7 y^2+5 y+\underline{\text{ }}\right)=7 \left(y^2+\frac{5 y}{7}+\underline{\text{ }}\right): \\
9 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$7 \left(y^2+\frac{5 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{9}{4}=\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{4}-2=\frac{1}{4}: \\
9 \left(x^2+x+\frac{1}{4}\right)+7 \left(y^2+\frac{5 y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{1}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }7\times \frac{25}{196}=\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{4}+\frac{25}{28}=\frac{8}{7}: \\
9 \left(x^2+x+\frac{1}{4}\right)+7 \left(y^2+\frac{5 y}{7}+\frac{25}{196}\right)=\fbox{$\frac{8}{7}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
9 \fbox{$\left(x+\frac{1}{2}\right)^2$}+7 \left(y^2+\frac{5 y}{7}+\frac{25}{196}\right)=\frac{8}{7} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{7}+\frac{25}{196}=\left(y+\frac{5}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x+\frac{1}{2}\right)^2+7 \fbox{$\left(y+\frac{5}{14}\right)^2$}=\frac{8}{7} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2-4 x+8 y^2-3 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-3 y+9 x^2-4 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
8 y^2-3 y+9 x^2-4 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(9 x^2-4 x+\underline{\text{ }}\right)+\left(8 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2-4 x+\underline{\text{ }}\right)=9 \left(x^2-\frac{4 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2-\frac{4 x}{9}+\underline{\text{ }}\right)$}+\left(8 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-3 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right): \\
9 \left(x^2-\frac{4 x}{9}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{9}}{2}\right)^2=\frac{4}{81} \text{on }\text{the }\text{left }\text{and }9\times \frac{4}{81}=\frac{4}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5+\frac{4}{9}=\frac{49}{9}: \\
9 \left(x^2-\frac{4 x}{9}+\frac{4}{81}\right)+8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right)=\fbox{$\frac{49}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{8}}{2}\right)^2=\frac{9}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{9}{256}=\frac{9}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{49}{9}+\frac{9}{32}=\frac{1649}{288}: \\
9 \left(x^2-\frac{4 x}{9}+\frac{4}{81}\right)+8 \left(y^2-\frac{3 y}{8}+\frac{9}{256}\right)=\fbox{$\frac{1649}{288}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{4 x}{9}+\frac{4}{81}=\left(x-\frac{2}{9}\right)^2: \\
9 \fbox{$\left(x-\frac{2}{9}\right)^2$}+8 \left(y^2-\frac{3 y}{8}+\frac{9}{256}\right)=\frac{1649}{288} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{8}+\frac{9}{256}=\left(y-\frac{3}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x-\frac{2}{9}\right)^2+8 \fbox{$\left(y-\frac{3}{16}\right)^2$}=\frac{1649}{288} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-10 x-2 y^2+y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+y-8 x^2-10 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-2 y^2+y-8 x^2-10 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2-10 x+\underline{\text{ }}\right)+\left(-2 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2-10 x+\underline{\text{ }}\right)=-8 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right)$}+\left(-2 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+y+\underline{\text{ }}\right)=-2 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right): \\
-8 \left(x^2+\frac{5 x}{4}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }-8\times \frac{25}{64}=-\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9-\frac{25}{8}=\frac{47}{8}: \\
-8 \left(x^2+\frac{5 x}{4}+\frac{25}{64}\right)-2 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{47}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-2}{16}=-\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{47}{8}-\frac{1}{8}=\frac{23}{4}: \\
-8 \left(x^2+\frac{5 x}{4}+\frac{25}{64}\right)-2 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{23}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{4}+\frac{25}{64}=\left(x+\frac{5}{8}\right)^2: \\
-8 \fbox{$\left(x+\frac{5}{8}\right)^2$}-2 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\frac{23}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{2}+\frac{1}{16}=\left(y-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x+\frac{5}{8}\right)^2-2 \fbox{$\left(y-\frac{1}{4}\right)^2$}=\frac{23}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2+x+10 y^2-2 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2-2 y+5 x^2+x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
10 y^2-2 y+5 x^2+x=7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2+x+\underline{\text{ }}\right)+\left(10 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+x+\underline{\text{ }}\right)=5 \left(x^2+\frac{x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+\frac{x}{5}+\underline{\text{ }}\right)$}+\left(10 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2-2 y+\underline{\text{ }}\right)=10 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right): \\
5 \left(x^2+\frac{x}{5}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{5}{100}=\frac{1}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
7+\frac{1}{20}=\frac{141}{20}: \\
5 \left(x^2+\frac{x}{5}+\frac{1}{100}\right)+10 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{141}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{10}{100}=\frac{1}{10} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{141}{20}+\frac{1}{10}=\frac{143}{20}: \\
5 \left(x^2+\frac{x}{5}+\frac{1}{100}\right)+10 \left(y^2-\frac{y}{5}+\frac{1}{100}\right)=\fbox{$\frac{143}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{5}+\frac{1}{100}=\left(x+\frac{1}{10}\right)^2: \\
5 \fbox{$\left(x+\frac{1}{10}\right)^2$}+10 \left(y^2-\frac{y}{5}+\frac{1}{100}\right)=\frac{143}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{5}+\frac{1}{100}=\left(y-\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x+\frac{1}{10}\right)^2+\text{10 }\fbox{$\left(y-\frac{1}{10}\right)^2$}=\frac{143}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2-5 x-6 y^2+9 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+9 y+9 x^2-5 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-6 y^2+9 y+9 x^2-5 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(9 x^2-5 x+\underline{\text{ }}\right)+\left(-6 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2-5 x+\underline{\text{ }}\right)=9 \left(x^2-\frac{5 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2-\frac{5 x}{9}+\underline{\text{ }}\right)$}+\left(-6 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2+9 y+\underline{\text{ }}\right)=-6 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right): \\
9 \left(x^2-\frac{5 x}{9}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{9}}{2}\right)^2=\frac{25}{324} \text{on }\text{the }\text{left }\text{and }9\times \frac{25}{324}=\frac{25}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+\frac{25}{36}=\frac{241}{36}: \\
9 \left(x^2-\frac{5 x}{9}+\frac{25}{324}\right)-6 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{241}{36}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }-6\times \frac{9}{16}=-\frac{27}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{241}{36}-\frac{27}{8}=\frac{239}{72}: \\
9 \left(x^2-\frac{5 x}{9}+\frac{25}{324}\right)-6 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$\frac{239}{72}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{9}+\frac{25}{324}=\left(x-\frac{5}{18}\right)^2: \\
9 \fbox{$\left(x-\frac{5}{18}\right)^2$}-6 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\frac{239}{72} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{2}+\frac{9}{16}=\left(y-\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x-\frac{5}{18}\right)^2-6 \fbox{$\left(y-\frac{3}{4}\right)^2$}=\frac{239}{72} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-4 x+8 y^2+3 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2+3 y-8 x^2-4 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
8 y^2+3 y-8 x^2-4 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2-4 x+\underline{\text{ }}\right)+\left(8 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2-4 x+\underline{\text{ }}\right)=-8 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)$}+\left(8 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2+3 y+\underline{\text{ }}\right)=8 \left(y^2+\frac{3 y}{8}+\underline{\text{ }}\right): \\
-8 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2+\frac{3 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-8}{16}=-\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6-\frac{1}{2}=\frac{11}{2}: \\
-8 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+8 \left(y^2+\frac{3 y}{8}+\underline{\text{ }}\right)=\fbox{$\frac{11}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{8}}{2}\right)^2=\frac{9}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{9}{256}=\frac{9}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{11}{2}+\frac{9}{32}=\frac{185}{32}: \\
-8 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+8 \left(y^2+\frac{3 y}{8}+\frac{9}{256}\right)=\fbox{$\frac{185}{32}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{2}+\frac{1}{16}=\left(x+\frac{1}{4}\right)^2: \\
-8 \fbox{$\left(x+\frac{1}{4}\right)^2$}+8 \left(y^2+\frac{3 y}{8}+\frac{9}{256}\right)=\frac{185}{32} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{8}+\frac{9}{256}=\left(y+\frac{3}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x+\frac{1}{4}\right)^2+8 \fbox{$\left(y+\frac{3}{16}\right)^2$}=\frac{185}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2+5 x+4 y^2+8 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 y^2+8 y-7 x^2+5 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
4 y^2+8 y-7 x^2+5 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-7 x^2+5 x+\underline{\text{ }}\right)+\left(4 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2+5 x+\underline{\text{ }}\right)=-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right)$}+\left(4 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 y^2+8 y+\underline{\text{ }}\right)=4 \left(y^2+2 y+\underline{\text{ }}\right): \\
-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right)+\fbox{$4 \left(y^2+2 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }-7\times \frac{25}{196}=-\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-4-\frac{25}{28}=-\frac{137}{28}: \\
-7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)+4 \left(y^2+2 y+\underline{\text{ }}\right)=\fbox{$-\frac{137}{28}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }4\times 1=4 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
4-\frac{137}{28}=-\frac{25}{28}: \\
-7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)+4 \left(y^2+2 y+1\right)=\fbox{$-\frac{25}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{7}+\frac{25}{196}=\left(x-\frac{5}{14}\right)^2: \\
-7 \fbox{$\left(x-\frac{5}{14}\right)^2$}+4 \left(y^2+2 y+1\right)=-\frac{25}{28} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x-\frac{5}{14}\right)^2+4 \fbox{$(y+1)^2$}=-\frac{25}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2-6 x+8 y^2+y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2+y+10 x^2-6 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
8 y^2+y+10 x^2-6 x=-3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(10 x^2-6 x+\underline{\text{ }}\right)+\left(8 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(10 x^2-6 x+\underline{\text{ }}\right)=10 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right)$}+\left(8 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2+y+\underline{\text{ }}\right)=8 \left(y^2+\frac{y}{8}+\underline{\text{ }}\right): \\
10 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2+\frac{y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}-3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }10\times \frac{9}{100}=\frac{9}{10} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{10}-3=-\frac{21}{10}: \\
10 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)+8 \left(y^2+\frac{y}{8}+\underline{\text{ }}\right)=\fbox{$-\frac{21}{10}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{8}}{2}\right)^2=\frac{1}{256} \text{on }\text{the }\text{left }\text{and }\frac{8}{256}=\frac{1}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{32}-\frac{21}{10}=-\frac{331}{160}: \\
10 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)+8 \left(y^2+\frac{y}{8}+\frac{1}{256}\right)=\fbox{$-\frac{331}{160}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{5}+\frac{9}{100}=\left(x-\frac{3}{10}\right)^2: \\
\text{10 }\fbox{$\left(x-\frac{3}{10}\right)^2$}+8 \left(y^2+\frac{y}{8}+\frac{1}{256}\right)=-\frac{331}{160} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{8}+\frac{1}{256}=\left(y+\frac{1}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 10 \left(x-\frac{3}{10}\right)^2+8 \fbox{$\left(y+\frac{1}{16}\right)^2$}=-\frac{331}{160} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2+5 x-y^2-3 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2-3 y-7 x^2+5 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-y^2-3 y-7 x^2+5 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-7 x^2+5 x+\underline{\text{ }}\right)+\left(-y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2+5 x+\underline{\text{ }}\right)=-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right)$}+\left(-y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2-3 y+\underline{\text{ }}\right)=-\left(y^2+3 y+\underline{\text{ }}\right): \\
-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right)+\fbox{$-\left(y^2+3 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }-7\times \frac{25}{196}=-\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-9-\frac{25}{28}=-\frac{277}{28}: \\
-7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)-\left(y^2+3 y+\underline{\text{ }}\right)=\fbox{$-\frac{277}{28}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }-\frac{9}{4}=-\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{277}{28}-\frac{9}{4}=-\frac{85}{7}: \\
-7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)-\left(y^2+3 y+\frac{9}{4}\right)=\fbox{$-\frac{85}{7}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{7}+\frac{25}{196}=\left(x-\frac{5}{14}\right)^2: \\
-7 \fbox{$\left(x-\frac{5}{14}\right)^2$}-\left(y^2+3 y+\frac{9}{4}\right)=-\frac{85}{7} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+3 y+\frac{9}{4}=\left(y+\frac{3}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x-\frac{5}{14}\right)^2-\fbox{$\left(y+\frac{3}{2}\right)^2$}=-\frac{85}{7} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-2 x+10 y^2+9 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2+9 y-8 x^2-2 x=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2-2 x+\underline{\text{ }}\right)+\left(10 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(-8 x^2-2 x+\underline{\text{ }}\right)=-8 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right)$}+\left(10 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(10 y^2+9 y+\underline{\text{ }}\right)=10 \left(y^2+\frac{9 y}{10}+\underline{\text{ }}\right): \\
-8 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2+\frac{9 y}{10}+\underline{\text{ }}\right)$}=\underline{\text{ }}+0 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{-8}{64}=-\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{10}}{2}\right)^2=\frac{81}{400} \text{on }\text{the }\text{left }\text{and }10\times \frac{81}{400}=\frac{81}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{81}{40}-\frac{1}{8}=\frac{19}{10}: \\
-8 \left(x^2+\frac{x}{4}+\frac{1}{64}\right)+10 \left(y^2+\frac{9 y}{10}+\frac{81}{400}\right)=\fbox{$\frac{19}{10}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{x}{4}+\frac{1}{64}=\left(x+\frac{1}{8}\right)^2: \\
-8 \fbox{$\left(x+\frac{1}{8}\right)^2$}+10 \left(y^2+\frac{9 y}{10}+\frac{81}{400}\right)=\frac{19}{10} \\
\end{array}
Step 9:
\begin{array}{l}
y^2+\frac{9 y}{10}+\frac{81}{400}=\left(y+\frac{9}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x+\frac{1}{8}\right)^2+\text{10 }\fbox{$\left(y+\frac{9}{20}\right)^2$}=\frac{19}{10} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-6 x+9 y^2-7 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2-7 y+2 x^2-6 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
9 y^2-7 y+2 x^2-6 x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2-6 x+\underline{\text{ }}\right)+\left(9 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-6 x+\underline{\text{ }}\right)=2 \left(x^2-3 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-3 x+\underline{\text{ }}\right)$}+\left(9 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2-7 y+\underline{\text{ }}\right)=9 \left(y^2-\frac{7 y}{9}+\underline{\text{ }}\right): \\
2 \left(x^2-3 x+\underline{\text{ }}\right)+\fbox{$9 \left(y^2-\frac{7 y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }2\times \frac{9}{4}=\frac{9}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2+\frac{9}{2}=\frac{13}{2}: \\
2 \left(x^2-3 x+\frac{9}{4}\right)+9 \left(y^2-\frac{7 y}{9}+\underline{\text{ }}\right)=\fbox{$\frac{13}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{9}}{2}\right)^2=\frac{49}{324} \text{on }\text{the }\text{left }\text{and }9\times \frac{49}{324}=\frac{49}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{13}{2}+\frac{49}{36}=\frac{283}{36}: \\
2 \left(x^2-3 x+\frac{9}{4}\right)+9 \left(y^2-\frac{7 y}{9}+\frac{49}{324}\right)=\fbox{$\frac{283}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-3 x+\frac{9}{4}=\left(x-\frac{3}{2}\right)^2: \\
2 \fbox{$\left(x-\frac{3}{2}\right)^2$}+9 \left(y^2-\frac{7 y}{9}+\frac{49}{324}\right)=\frac{283}{36} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{9}+\frac{49}{324}=\left(y-\frac{7}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x-\frac{3}{2}\right)^2+9 \fbox{$\left(y-\frac{7}{18}\right)^2$}=\frac{283}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-9 x+2 y^2-7 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-7 y+x^2-9 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
2 y^2-7 y+x^2-9 x=4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(x^2-9 x+\underline{\text{ }}\right)+\left(2 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 y^2-7 y+\underline{\text{ }}\right)=2 \left(y^2-\frac{7 y}{2}+\underline{\text{ }}\right): \\
\left(x^2-9 x+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-\frac{7 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-9}{2}\right)^2=\frac{81}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
4+\frac{81}{4}=\frac{97}{4}: \\
\left(x^2-9 x+\frac{81}{4}\right)+2 \left(y^2-\frac{7 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{97}{4}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{2}}{2}\right)^2=\frac{49}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{49}{16}=\frac{49}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{97}{4}+\frac{49}{8}=\frac{243}{8}: \\
\left(x^2-9 x+\frac{81}{4}\right)+2 \left(y^2-\frac{7 y}{2}+\frac{49}{16}\right)=\fbox{$\frac{243}{8}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-9 x+\frac{81}{4}=\left(x-\frac{9}{2}\right)^2: \\
\fbox{$\left(x-\frac{9}{2}\right)^2$}+2 \left(y^2-\frac{7 y}{2}+\frac{49}{16}\right)=\frac{243}{8} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-\frac{7 y}{2}+\frac{49}{16}=\left(y-\frac{7}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \left(x-\frac{9}{2}\right)^2+2 \fbox{$\left(y-\frac{7}{4}\right)^2$}=\frac{243}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2+9 x-10 y^2+8 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+8 y+3 x^2+9 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
-10 y^2+8 y+3 x^2+9 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2+9 x+\underline{\text{ }}\right)+\left(-10 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2+9 x+\underline{\text{ }}\right)=3 \left(x^2+3 x+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2+3 x+\underline{\text{ }}\right)$}+\left(-10 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2+8 y+\underline{\text{ }}\right)=-10 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right): \\
3 \left(x^2+3 x+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }3\times \frac{9}{4}=\frac{27}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3+\frac{27}{4}=\frac{39}{4}: \\
3 \left(x^2+3 x+\frac{9}{4}\right)-10 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{39}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }-10\times \frac{4}{25}=-\frac{8}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{39}{4}-\frac{8}{5}=\frac{163}{20}: \\
3 \left(x^2+3 x+\frac{9}{4}\right)-10 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=\fbox{$\frac{163}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+3 x+\frac{9}{4}=\left(x+\frac{3}{2}\right)^2: \\
3 \fbox{$\left(x+\frac{3}{2}\right)^2$}-10 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=\frac{163}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{4 y}{5}+\frac{4}{25}=\left(y-\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x+\frac{3}{2}\right)^2-\text{10 }\fbox{$\left(y-\frac{2}{5}\right)^2$}=\frac{163}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2-6 x-6 y^2+9 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+9 y+8 x^2-6 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-6 y^2+9 y+8 x^2-6 x=8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(8 x^2-6 x+\underline{\text{ }}\right)+\left(-6 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2-6 x+\underline{\text{ }}\right)=8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)$}+\left(-6 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2+9 y+\underline{\text{ }}\right)=-6 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right): \\
8 \left(x^2-\frac{3 x}{4}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }8\times \frac{9}{64}=\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8+\frac{9}{8}=\frac{73}{8}: \\
8 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)-6 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{73}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }-6\times \frac{9}{16}=-\frac{27}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{73}{8}-\frac{27}{8}=\frac{23}{4}: \\
8 \left(x^2-\frac{3 x}{4}+\frac{9}{64}\right)-6 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$\frac{23}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{4}+\frac{9}{64}=\left(x-\frac{3}{8}\right)^2: \\
8 \fbox{$\left(x-\frac{3}{8}\right)^2$}-6 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\frac{23}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{2}+\frac{9}{16}=\left(y-\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x-\frac{3}{8}\right)^2-6 \fbox{$\left(y-\frac{3}{4}\right)^2$}=\frac{23}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $8 x^2+8 x-10 y^2-7 y+6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-7 y+8 x^2+8 x+6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 \text{from }\text{both }\text{sides}: \\
-10 y^2-7 y+8 x^2+8 x=-6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(8 x^2+8 x+\underline{\text{ }}\right)+\left(-10 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(8 x^2+8 x+\underline{\text{ }}\right)=8 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$8 \left(x^2+x+\underline{\text{ }}\right)$}+\left(-10 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2-7 y+\underline{\text{ }}\right)=-10 \left(y^2+\frac{7 y}{10}+\underline{\text{ }}\right): \\
8 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+\frac{7 y}{10}+\underline{\text{ }}\right)$}=\underline{\text{ }}-6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{8}{4}=2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2-6=-4: \\
8 \left(x^2+x+\frac{1}{4}\right)-10 \left(y^2+\frac{7 y}{10}+\underline{\text{ }}\right)=\fbox{$-4$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{10}}{2}\right)^2=\frac{49}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{49}{400}=-\frac{49}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-4-\frac{49}{40}=-\frac{209}{40}: \\
8 \left(x^2+x+\frac{1}{4}\right)-10 \left(y^2+\frac{7 y}{10}+\frac{49}{400}\right)=\fbox{$-\frac{209}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
8 \fbox{$\left(x+\frac{1}{2}\right)^2$}-10 \left(y^2+\frac{7 y}{10}+\frac{49}{400}\right)=-\frac{209}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{7 y}{10}+\frac{49}{400}=\left(y+\frac{7}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 8 \left(x+\frac{1}{2}\right)^2-\text{10 }\fbox{$\left(y+\frac{7}{20}\right)^2$}=-\frac{209}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-10 x-10 y^2-3 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-3 y+4 x^2-10 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-10 y^2-3 y+4 x^2-10 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2-10 x+\underline{\text{ }}\right)+\left(-10 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2-10 x+\underline{\text{ }}\right)=4 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)$}+\left(-10 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2-3 y+\underline{\text{ }}\right)=-10 \left(y^2+\frac{3 y}{10}+\underline{\text{ }}\right): \\
4 \left(x^2-\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+\frac{3 y}{10}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }4\times \frac{25}{16}=\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9+\frac{25}{4}=\frac{61}{4}: \\
4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)-10 \left(y^2+\frac{3 y}{10}+\underline{\text{ }}\right)=\fbox{$\frac{61}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{10}}{2}\right)^2=\frac{9}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{9}{400}=-\frac{9}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{61}{4}-\frac{9}{40}=\frac{601}{40}: \\
4 \left(x^2-\frac{5 x}{2}+\frac{25}{16}\right)-10 \left(y^2+\frac{3 y}{10}+\frac{9}{400}\right)=\fbox{$\frac{601}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{2}+\frac{25}{16}=\left(x-\frac{5}{4}\right)^2: \\
4 \fbox{$\left(x-\frac{5}{4}\right)^2$}-10 \left(y^2+\frac{3 y}{10}+\frac{9}{400}\right)=\frac{601}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{10}+\frac{9}{400}=\left(y+\frac{3}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x-\frac{5}{4}\right)^2-\text{10 }\fbox{$\left(y+\frac{3}{20}\right)^2$}=\frac{601}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-9 x-y^2-7 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2-7 y-6 x^2-9 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-y^2-7 y-6 x^2-9 x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2-9 x+\underline{\text{ }}\right)+\left(-y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2-9 x+\underline{\text{ }}\right)=-6 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right)$}+\left(-y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2-7 y+\underline{\text{ }}\right)=-\left(y^2+7 y+\underline{\text{ }}\right): \\
-6 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right)+\fbox{$-\left(y^2+7 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }-6\times \frac{9}{16}=-\frac{27}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-7-\frac{27}{8}=-\frac{83}{8}: \\
-6 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)-\left(y^2+7 y+\underline{\text{ }}\right)=\fbox{$-\frac{83}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{7}{2}\right)^2=\frac{49}{4} \text{on }\text{the }\text{left }\text{and }-\frac{49}{4}=-\frac{49}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{83}{8}-\frac{49}{4}=-\frac{181}{8}: \\
-6 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)-\left(y^2+7 y+\frac{49}{4}\right)=\fbox{$-\frac{181}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{2}+\frac{9}{16}=\left(x+\frac{3}{4}\right)^2: \\
-6 \fbox{$\left(x+\frac{3}{4}\right)^2$}-\left(y^2+7 y+\frac{49}{4}\right)=-\frac{181}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+7 y+\frac{49}{4}=\left(y+\frac{7}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x+\frac{3}{4}\right)^2-\fbox{$\left(y+\frac{7}{2}\right)^2$}=-\frac{181}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2-3 x-9 y^2-9 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-9 y^2-9 y-2 x^2-3 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
-9 y^2-9 y-2 x^2-3 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-2 x^2-3 x+\underline{\text{ }}\right)+\left(-9 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 x^2-3 x+\underline{\text{ }}\right)=-2 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right): \\
\fbox{$-2 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right)$}+\left(-9 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-9 y^2-9 y+\underline{\text{ }}\right)=-9 \left(y^2+y+\underline{\text{ }}\right): \\
-2 \left(x^2+\frac{3 x}{2}+\underline{\text{ }}\right)+\fbox{$-9 \left(y^2+y+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }-2\times \frac{9}{16}=-\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3-\frac{9}{8}=\frac{15}{8}: \\
-2 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)-9 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$\frac{15}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-9}{4}=-\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{15}{8}-\frac{9}{4}=-\frac{3}{8}: \\
-2 \left(x^2+\frac{3 x}{2}+\frac{9}{16}\right)-9 \left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{3}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{2}+\frac{9}{16}=\left(x+\frac{3}{4}\right)^2: \\
-2 \fbox{$\left(x+\frac{3}{4}\right)^2$}-9 \left(y^2+y+\frac{1}{4}\right)=-\frac{3}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -2 \left(x+\frac{3}{4}\right)^2-9 \fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{3}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+9 x+2 y^2-3 y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-3 y-8 x^2+9 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
2 y^2-3 y-8 x^2+9 x=8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2+9 x+\underline{\text{ }}\right)+\left(2 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2+9 x+\underline{\text{ }}\right)=-8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right)$}+\left(2 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2-3 y+\underline{\text{ }}\right)=2 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{8}}{2}\right)^2=\frac{81}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{81}{256}=-\frac{81}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8-\frac{81}{32}=\frac{175}{32}: \\
-8 \left(x^2-\frac{9 x}{8}+\frac{81}{256}\right)+2 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{175}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{9}{16}=\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{175}{32}+\frac{9}{8}=\frac{211}{32}: \\
-8 \left(x^2-\frac{9 x}{8}+\frac{81}{256}\right)+2 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$\frac{211}{32}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{9 x}{8}+\frac{81}{256}=\left(x-\frac{9}{16}\right)^2: \\
-8 \fbox{$\left(x-\frac{9}{16}\right)^2$}+2 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\frac{211}{32} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{2}+\frac{9}{16}=\left(y-\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x-\frac{9}{16}\right)^2+2 \fbox{$\left(y-\frac{3}{4}\right)^2$}=\frac{211}{32} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2-8 x+5 y^2-y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2-y-4 x^2-8 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
5 y^2-y-4 x^2-8 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-4 x^2-8 x+\underline{\text{ }}\right)+\left(5 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2-8 x+\underline{\text{ }}\right)=-4 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2+2 x+\underline{\text{ }}\right)$}+\left(5 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2-y+\underline{\text{ }}\right)=5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right): \\
-4 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-4\times 1=-4 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-1-4=-5: \\
-4 \left(x^2+2 x+1\right)+5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right)=\fbox{$-5$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{5}{100}=\frac{1}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{20}-5=-\frac{99}{20}: \\
-4 \left(x^2+2 x+1\right)+5 \left(y^2-\frac{y}{5}+\frac{1}{100}\right)=\fbox{$-\frac{99}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
-4 \fbox{$(x+1)^2$}+5 \left(y^2-\frac{y}{5}+\frac{1}{100}\right)=-\frac{99}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{5}+\frac{1}{100}=\left(y-\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 (x+1)^2+5 \fbox{$\left(y-\frac{1}{10}\right)^2$}=-\frac{99}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2+7 x+6 y^2+9 y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+9 y+9 x^2+7 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
6 y^2+9 y+9 x^2+7 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(9 x^2+7 x+\underline{\text{ }}\right)+\left(6 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2+7 x+\underline{\text{ }}\right)=9 \left(x^2+\frac{7 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2+\frac{7 x}{9}+\underline{\text{ }}\right)$}+\left(6 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2+9 y+\underline{\text{ }}\right)=6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right): \\
9 \left(x^2+\frac{7 x}{9}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{9}}{2}\right)^2=\frac{49}{324} \text{on }\text{the }\text{left }\text{and }9\times \frac{49}{324}=\frac{49}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5+\frac{49}{36}=\frac{229}{36}: \\
9 \left(x^2+\frac{7 x}{9}+\frac{49}{324}\right)+6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{229}{36}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }6\times \frac{9}{16}=\frac{27}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{229}{36}+\frac{27}{8}=\frac{701}{72}: \\
9 \left(x^2+\frac{7 x}{9}+\frac{49}{324}\right)+6 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$\frac{701}{72}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{9}+\frac{49}{324}=\left(x+\frac{7}{18}\right)^2: \\
9 \fbox{$\left(x+\frac{7}{18}\right)^2$}+6 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\frac{701}{72} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{2}+\frac{9}{16}=\left(y+\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x+\frac{7}{18}\right)^2+6 \fbox{$\left(y+\frac{3}{4}\right)^2$}=\frac{701}{72} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2-10 x+6 y^2+9 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2+9 y+9 x^2-10 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
6 y^2+9 y+9 x^2-10 x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(9 x^2-10 x+\underline{\text{ }}\right)+\left(6 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2-10 x+\underline{\text{ }}\right)=9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right)$}+\left(6 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2+9 y+\underline{\text{ }}\right)=6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right): \\
9 \left(x^2-\frac{10 x}{9}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-10}{9}}{2}\right)^2=\frac{25}{81} \text{on }\text{the }\text{left }\text{and }9\times \frac{25}{81}=\frac{25}{9} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2+\frac{25}{9}=\frac{43}{9}: \\
9 \left(x^2-\frac{10 x}{9}+\frac{25}{81}\right)+6 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{43}{9}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }6\times \frac{9}{16}=\frac{27}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{43}{9}+\frac{27}{8}=\frac{587}{72}: \\
9 \left(x^2-\frac{10 x}{9}+\frac{25}{81}\right)+6 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$\frac{587}{72}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{10 x}{9}+\frac{25}{81}=\left(x-\frac{5}{9}\right)^2: \\
9 \fbox{$\left(x-\frac{5}{9}\right)^2$}+6 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\frac{587}{72} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{2}+\frac{9}{16}=\left(y+\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x-\frac{5}{9}\right)^2+6 \fbox{$\left(y+\frac{3}{4}\right)^2$}=\frac{587}{72} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2+7 x+6 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 x^2+7 x+(6 y-3)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }6 y-4 x^2+7 x-3 \text{from }\text{both }\text{sides}: \\
4 x^2-7 x+(3-6 y)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }3-6 y \text{from }\text{both }\text{sides}: \\
4 x^2-7 x=6 y-3 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(4 x^2-7 x+\underline{\text{ }}\right)=(6 y-3)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 x^2-7 x+\underline{\text{ }}\right)=4 \left(x^2-\frac{7 x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-\frac{7 x}{4}+\underline{\text{ }}\right)$}=(6 y-3)+\underline{\text{ }} \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{4}}{2}\right)^2=\frac{49}{64} \text{on }\text{the }\text{left }\text{and }4\times \frac{49}{64}=\frac{49}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
(6 y-3)+\frac{49}{16}=6 y+\frac{1}{16}: \\
4 \left(x^2-\frac{7 x}{4}+\frac{49}{64}\right)=\fbox{$6 y+\frac{1}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{7 x}{4}+\frac{49}{64}=\left(x-\frac{7}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \fbox{$\left(x-\frac{7}{8}\right)^2$}=6 y+\frac{1}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-x+6 y^2-y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2-y-3 x^2-x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
6 y^2-y-3 x^2-x=-3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2-x+\underline{\text{ }}\right)+\left(6 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-x+\underline{\text{ }}\right)=-3 \left(x^2+\frac{x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+\frac{x}{3}+\underline{\text{ }}\right)$}+\left(6 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2-y+\underline{\text{ }}\right)=6 \left(y^2-\frac{y}{6}+\underline{\text{ }}\right): \\
-3 \left(x^2+\frac{x}{3}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2-\frac{y}{6}+\underline{\text{ }}\right)$}=\underline{\text{ }}-3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{-3}{36}=-\frac{1}{12} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-3-\frac{1}{12}=-\frac{37}{12}: \\
-3 \left(x^2+\frac{x}{3}+\frac{1}{36}\right)+6 \left(y^2-\frac{y}{6}+\underline{\text{ }}\right)=\fbox{$-\frac{37}{12}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{6}}{2}\right)^2=\frac{1}{144} \text{on }\text{the }\text{left }\text{and }\frac{6}{144}=\frac{1}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{24}-\frac{37}{12}=-\frac{73}{24}: \\
-3 \left(x^2+\frac{x}{3}+\frac{1}{36}\right)+6 \left(y^2-\frac{y}{6}+\frac{1}{144}\right)=\fbox{$-\frac{73}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{3}+\frac{1}{36}=\left(x+\frac{1}{6}\right)^2: \\
-3 \fbox{$\left(x+\frac{1}{6}\right)^2$}+6 \left(y^2-\frac{y}{6}+\frac{1}{144}\right)=-\frac{73}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{6}+\frac{1}{144}=\left(y-\frac{1}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x+\frac{1}{6}\right)^2+6 \fbox{$\left(y-\frac{1}{12}\right)^2$}=-\frac{73}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2+7 x+8 y^2-4 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-4 y+10 x^2+7 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
8 y^2-4 y+10 x^2+7 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(10 x^2+7 x+\underline{\text{ }}\right)+\left(8 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(10 x^2+7 x+\underline{\text{ }}\right)=10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right)$}+\left(8 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-4 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right): \\
10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{10}}{2}\right)^2=\frac{49}{400} \text{on }\text{the }\text{left }\text{and }10\times \frac{49}{400}=\frac{49}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10+\frac{49}{40}=\frac{449}{40}: \\
10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)+8 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{449}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{8}{16}=\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{449}{40}+\frac{1}{2}=\frac{469}{40}: \\
10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)+8 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{469}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{10}+\frac{49}{400}=\left(x+\frac{7}{20}\right)^2: \\
\text{10 }\fbox{$\left(x+\frac{7}{20}\right)^2$}+8 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\frac{469}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{2}+\frac{1}{16}=\left(y-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 10 \left(x+\frac{7}{20}\right)^2+8 \fbox{$\left(y-\frac{1}{4}\right)^2$}=\frac{469}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-9 x+y^2-8 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2-8 y+5 x^2-9 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
y^2-8 y+5 x^2-9 x=-8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2-9 x+\underline{\text{ }}\right)+\left(y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2-9 x+\underline{\text{ }}\right)=5 \left(x^2-\frac{9 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-\frac{9 x}{5}+\underline{\text{ }}\right)$}+\left(y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{5}}{2}\right)^2=\frac{81}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{81}{100}=\frac{81}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\frac{81}{20}-8=-\frac{79}{20}: \\
5 \left(x^2-\frac{9 x}{5}+\frac{81}{100}\right)+\left(y^2-8 y+\underline{\text{ }}\right)=\fbox{$-\frac{79}{20}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-8}{2}\right)^2=16 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
16-\frac{79}{20}=\frac{241}{20}: \\
5 \left(x^2-\frac{9 x}{5}+\frac{81}{100}\right)+\left(y^2-8 y+16\right)=\fbox{$\frac{241}{20}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-\frac{9 x}{5}+\frac{81}{100}=\left(x-\frac{9}{10}\right)^2: \\
5 \fbox{$\left(x-\frac{9}{10}\right)^2$}+\left(y^2-8 y+16\right)=\frac{241}{20} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-8 y+16=(y-4)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x-\frac{9}{10}\right)^2+\fbox{$(y-4)^2$}=\frac{241}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2-x+5 y^2+10 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+10 y-7 x^2-x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
5 y^2+10 y-7 x^2-x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-7 x^2-x+\underline{\text{ }}\right)+\left(5 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2-x+\underline{\text{ }}\right)=-7 \left(x^2+\frac{x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2+\frac{x}{7}+\underline{\text{ }}\right)$}+\left(5 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+10 y+\underline{\text{ }}\right)=5 \left(y^2+2 y+\underline{\text{ }}\right): \\
-7 \left(x^2+\frac{x}{7}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+2 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{7}}{2}\right)^2=\frac{1}{196} \text{on }\text{the }\text{left }\text{and }\frac{-7}{196}=-\frac{1}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2-\frac{1}{28}=\frac{55}{28}: \\
-7 \left(x^2+\frac{x}{7}+\frac{1}{196}\right)+5 \left(y^2+2 y+\underline{\text{ }}\right)=\fbox{$\frac{55}{28}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }5\times 1=5 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{55}{28}+5=\frac{195}{28}: \\
-7 \left(x^2+\frac{x}{7}+\frac{1}{196}\right)+5 \left(y^2+2 y+1\right)=\fbox{$\frac{195}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{7}+\frac{1}{196}=\left(x+\frac{1}{14}\right)^2: \\
-7 \fbox{$\left(x+\frac{1}{14}\right)^2$}+5 \left(y^2+2 y+1\right)=\frac{195}{28} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x+\frac{1}{14}\right)^2+5 \fbox{$(y+1)^2$}=\frac{195}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2-3 x+8 y^2-8 y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-8 y+3 x^2-3 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
8 y^2-8 y+3 x^2-3 x=-3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2-3 x+\underline{\text{ }}\right)+\left(8 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2-3 x+\underline{\text{ }}\right)=3 \left(x^2-x+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2-x+\underline{\text{ }}\right)$}+\left(8 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-8 y+\underline{\text{ }}\right)=8 \left(y^2-y+\underline{\text{ }}\right): \\
3 \left(x^2-x+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}-3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{3}{4}=\frac{3}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{3}{4}-3=-\frac{9}{4}: \\
3 \left(x^2-x+\frac{1}{4}\right)+8 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$-\frac{9}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{8}{4}=2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
2-\frac{9}{4}=-\frac{1}{4}: \\
3 \left(x^2-x+\frac{1}{4}\right)+8 \left(y^2-y+\frac{1}{4}\right)=\fbox{$-\frac{1}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
3 \fbox{$\left(x-\frac{1}{2}\right)^2$}+8 \left(y^2-y+\frac{1}{4}\right)=-\frac{1}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x-\frac{1}{2}\right)^2+8 \fbox{$\left(y-\frac{1}{2}\right)^2$}=-\frac{1}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2+3 x-10 y^2+8 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+8 y+5 x^2+3 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-10 y^2+8 y+5 x^2+3 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2+3 x+\underline{\text{ }}\right)+\left(-10 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+3 x+\underline{\text{ }}\right)=5 \left(x^2+\frac{3 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+\frac{3 x}{5}+\underline{\text{ }}\right)$}+\left(-10 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2+8 y+\underline{\text{ }}\right)=-10 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right): \\
5 \left(x^2+\frac{3 x}{5}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{9}{100}=\frac{9}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{9}{20}-1=-\frac{11}{20}: \\
5 \left(x^2+\frac{3 x}{5}+\frac{9}{100}\right)-10 \left(y^2-\frac{4 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{11}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }-10\times \frac{4}{25}=-\frac{8}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{11}{20}-\frac{8}{5}=-\frac{43}{20}: \\
5 \left(x^2+\frac{3 x}{5}+\frac{9}{100}\right)-10 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=\fbox{$-\frac{43}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{3 x}{5}+\frac{9}{100}=\left(x+\frac{3}{10}\right)^2: \\
5 \fbox{$\left(x+\frac{3}{10}\right)^2$}-10 \left(y^2-\frac{4 y}{5}+\frac{4}{25}\right)=-\frac{43}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{4 y}{5}+\frac{4}{25}=\left(y-\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x+\frac{3}{10}\right)^2-\text{10 }\fbox{$\left(y-\frac{2}{5}\right)^2$}=-\frac{43}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+10 x+9 y^2+3 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+3 y+7 x^2+10 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
9 y^2+3 y+7 x^2+10 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(7 x^2+10 x+\underline{\text{ }}\right)+\left(9 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2+10 x+\underline{\text{ }}\right)=7 \left(x^2+\frac{10 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+\frac{10 x}{7}+\underline{\text{ }}\right)$}+\left(9 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2+3 y+\underline{\text{ }}\right)=9 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right): \\
7 \left(x^2+\frac{10 x}{7}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{10}{7}}{2}\right)^2=\frac{25}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{25}{49}=\frac{25}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+\frac{25}{7}=\frac{67}{7}: \\
7 \left(x^2+\frac{10 x}{7}+\frac{25}{49}\right)+9 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{67}{7}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{9}{36}=\frac{1}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{67}{7}+\frac{1}{4}=\frac{275}{28}: \\
7 \left(x^2+\frac{10 x}{7}+\frac{25}{49}\right)+9 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=\fbox{$\frac{275}{28}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{10 x}{7}+\frac{25}{49}=\left(x+\frac{5}{7}\right)^2: \\
7 \fbox{$\left(x+\frac{5}{7}\right)^2$}+9 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=\frac{275}{28} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{3}+\frac{1}{36}=\left(y+\frac{1}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x+\frac{5}{7}\right)^2+9 \fbox{$\left(y+\frac{1}{6}\right)^2$}=\frac{275}{28} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+x-3 y^2-y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2-y+4 x^2+x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-3 y^2-y+4 x^2+x=-8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2+x+\underline{\text{ }}\right)+\left(-3 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2+x+\underline{\text{ }}\right)=4 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right)$}+\left(-3 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2-y+\underline{\text{ }}\right)=-3 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right): \\
4 \left(x^2+\frac{x}{4}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}-8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{4}{64}=\frac{1}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{1}{16}-8=-\frac{127}{16}: \\
4 \left(x^2+\frac{x}{4}+\frac{1}{64}\right)-3 \left(y^2+\frac{y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{127}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{-3}{36}=-\frac{1}{12} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{127}{16}-\frac{1}{12}=-\frac{385}{48}: \\
4 \left(x^2+\frac{x}{4}+\frac{1}{64}\right)-3 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=\fbox{$-\frac{385}{48}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{4}+\frac{1}{64}=\left(x+\frac{1}{8}\right)^2: \\
4 \fbox{$\left(x+\frac{1}{8}\right)^2$}-3 \left(y^2+\frac{y}{3}+\frac{1}{36}\right)=-\frac{385}{48} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{3}+\frac{1}{36}=\left(y+\frac{1}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x+\frac{1}{8}\right)^2-3 \fbox{$\left(y+\frac{1}{6}\right)^2$}=-\frac{385}{48} \\
\end{array}
| khanacademy | amps |
Given the equation $3 x^2+8 x+2 y^2-8 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-8 y+3 x^2+8 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
2 y^2-8 y+3 x^2+8 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(3 x^2+8 x+\underline{\text{ }}\right)+\left(2 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(3 x^2+8 x+\underline{\text{ }}\right)=3 \left(x^2+\frac{8 x}{3}+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2+\frac{8 x}{3}+\underline{\text{ }}\right)$}+\left(2 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2-8 y+\underline{\text{ }}\right)=2 \left(y^2-4 y+\underline{\text{ }}\right): \\
3 \left(x^2+\frac{8 x}{3}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-4 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{3}}{2}\right)^2=\frac{16}{9} \text{on }\text{the }\text{left }\text{and }3\times \frac{16}{9}=\frac{16}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{16}{3}-10=-\frac{14}{3}: \\
3 \left(x^2+\frac{8 x}{3}+\frac{16}{9}\right)+2 \left(y^2-4 y+\underline{\text{ }}\right)=\fbox{$-\frac{14}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-4}{2}\right)^2=4 \text{on }\text{the }\text{left }\text{and }2\times 4=8 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
8-\frac{14}{3}=\frac{10}{3}: \\
3 \left(x^2+\frac{8 x}{3}+\frac{16}{9}\right)+2 \left(y^2-4 y+4\right)=\fbox{$\frac{10}{3}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{8 x}{3}+\frac{16}{9}=\left(x+\frac{4}{3}\right)^2: \\
3 \fbox{$\left(x+\frac{4}{3}\right)^2$}+2 \left(y^2-4 y+4\right)=\frac{10}{3} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-4 y+4=(y-2)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \left(x+\frac{4}{3}\right)^2+2 \fbox{$(y-2)^2$}=\frac{10}{3} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2+4 x+8 y^2-3 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-3 y+7 x^2+4 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
8 y^2-3 y+7 x^2+4 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(7 x^2+4 x+\underline{\text{ }}\right)+\left(8 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2+4 x+\underline{\text{ }}\right)=7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right)$}+\left(8 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-3 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right): \\
7 \left(x^2+\frac{4 x}{7}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{7}}{2}\right)^2=\frac{4}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{4}{49}=\frac{4}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10+\frac{4}{7}=\frac{74}{7}: \\
7 \left(x^2+\frac{4 x}{7}+\frac{4}{49}\right)+8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right)=\fbox{$\frac{74}{7}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{8}}{2}\right)^2=\frac{9}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{9}{256}=\frac{9}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{74}{7}+\frac{9}{32}=\frac{2431}{224}: \\
7 \left(x^2+\frac{4 x}{7}+\frac{4}{49}\right)+8 \left(y^2-\frac{3 y}{8}+\frac{9}{256}\right)=\fbox{$\frac{2431}{224}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{4 x}{7}+\frac{4}{49}=\left(x+\frac{2}{7}\right)^2: \\
7 \fbox{$\left(x+\frac{2}{7}\right)^2$}+8 \left(y^2-\frac{3 y}{8}+\frac{9}{256}\right)=\frac{2431}{224} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{8}+\frac{9}{256}=\left(y-\frac{3}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x+\frac{2}{7}\right)^2+8 \fbox{$\left(y-\frac{3}{16}\right)^2$}=\frac{2431}{224} \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2+2 x-y^2-5 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2-5 y-2 x^2+2 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-y^2-5 y-2 x^2+2 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-2 x^2+2 x+\underline{\text{ }}\right)+\left(-y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-2 x^2+2 x+\underline{\text{ }}\right)=-2 \left(x^2-x+\underline{\text{ }}\right): \\
\fbox{$-2 \left(x^2-x+\underline{\text{ }}\right)$}+\left(-y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2-5 y+\underline{\text{ }}\right)=-\left(y^2+5 y+\underline{\text{ }}\right): \\
-2 \left(x^2-x+\underline{\text{ }}\right)+\fbox{$-\left(y^2+5 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-2}{4}=-\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-1-\frac{1}{2}=-\frac{3}{2}: \\
-2 \left(x^2-x+\frac{1}{4}\right)-\left(y^2+5 y+\underline{\text{ }}\right)=\fbox{$-\frac{3}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }-\frac{25}{4}=-\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{3}{2}-\frac{25}{4}=-\frac{31}{4}: \\
-2 \left(x^2-x+\frac{1}{4}\right)-\left(y^2+5 y+\frac{25}{4}\right)=\fbox{$-\frac{31}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
-2 \fbox{$\left(x-\frac{1}{2}\right)^2$}-\left(y^2+5 y+\frac{25}{4}\right)=-\frac{31}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+5 y+\frac{25}{4}=\left(y+\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -2 \left(x-\frac{1}{2}\right)^2-\fbox{$\left(y+\frac{5}{2}\right)^2$}=-\frac{31}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2-7 x+2 y^2+10 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+10 y+9 x^2-7 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
2 y^2+10 y+9 x^2-7 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(9 x^2-7 x+\underline{\text{ }}\right)+\left(2 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2-7 x+\underline{\text{ }}\right)=9 \left(x^2-\frac{7 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2-\frac{7 x}{9}+\underline{\text{ }}\right)$}+\left(2 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+10 y+\underline{\text{ }}\right)=2 \left(y^2+5 y+\underline{\text{ }}\right): \\
9 \left(x^2-\frac{7 x}{9}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+5 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{9}}{2}\right)^2=\frac{49}{324} \text{on }\text{the }\text{left }\text{and }9\times \frac{49}{324}=\frac{49}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{49}{36}-4=-\frac{95}{36}: \\
9 \left(x^2-\frac{7 x}{9}+\frac{49}{324}\right)+2 \left(y^2+5 y+\underline{\text{ }}\right)=\fbox{$-\frac{95}{36}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{5}{2}\right)^2=\frac{25}{4} \text{on }\text{the }\text{left }\text{and }2\times \frac{25}{4}=\frac{25}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{2}-\frac{95}{36}=\frac{355}{36}: \\
9 \left(x^2-\frac{7 x}{9}+\frac{49}{324}\right)+2 \left(y^2+5 y+\frac{25}{4}\right)=\fbox{$\frac{355}{36}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{7 x}{9}+\frac{49}{324}=\left(x-\frac{7}{18}\right)^2: \\
9 \fbox{$\left(x-\frac{7}{18}\right)^2$}+2 \left(y^2+5 y+\frac{25}{4}\right)=\frac{355}{36} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+5 y+\frac{25}{4}=\left(y+\frac{5}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x-\frac{7}{18}\right)^2+2 \fbox{$\left(y+\frac{5}{2}\right)^2$}=\frac{355}{36} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2-10 x+9 y^2+y+3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
9 y^2+y-4 x^2-10 x+3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }3 \text{from }\text{both }\text{sides}: \\
9 y^2+y-4 x^2-10 x=-3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-4 x^2-10 x+\underline{\text{ }}\right)+\left(9 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2-10 x+\underline{\text{ }}\right)=-4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)$}+\left(9 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(9 y^2+y+\underline{\text{ }}\right)=9 \left(y^2+\frac{y}{9}+\underline{\text{ }}\right): \\
-4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$9 \left(y^2+\frac{y}{9}+\underline{\text{ }}\right)$}=\underline{\text{ }}-3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }-4\times \frac{25}{16}=-\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-3-\frac{25}{4}=-\frac{37}{4}: \\
-4 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)+9 \left(y^2+\frac{y}{9}+\underline{\text{ }}\right)=\fbox{$-\frac{37}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{9}}{2}\right)^2=\frac{1}{324} \text{on }\text{the }\text{left }\text{and }\frac{9}{324}=\frac{1}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{36}-\frac{37}{4}=-\frac{83}{9}: \\
-4 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)+9 \left(y^2+\frac{y}{9}+\frac{1}{324}\right)=\fbox{$-\frac{83}{9}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{2}+\frac{25}{16}=\left(x+\frac{5}{4}\right)^2: \\
-4 \fbox{$\left(x+\frac{5}{4}\right)^2$}+9 \left(y^2+\frac{y}{9}+\frac{1}{324}\right)=-\frac{83}{9} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{9}+\frac{1}{324}=\left(y+\frac{1}{18}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x+\frac{5}{4}\right)^2+9 \fbox{$\left(y+\frac{1}{18}\right)^2$}=-\frac{83}{9} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-4 x-7 y^2+4 y+8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2+4 y-6 x^2-4 x+8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }8 \text{from }\text{both }\text{sides}: \\
-7 y^2+4 y-6 x^2-4 x=-8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2-4 x+\underline{\text{ }}\right)+\left(-7 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2-4 x+\underline{\text{ }}\right)=-6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(-7 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2+4 y+\underline{\text{ }}\right)=-7 \left(y^2-\frac{4 y}{7}+\underline{\text{ }}\right): \\
-6 \left(x^2+\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2-\frac{4 y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}-8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{-6}{9}=-\frac{2}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-8-\frac{2}{3}=-\frac{26}{3}: \\
-6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)-7 \left(y^2-\frac{4 y}{7}+\underline{\text{ }}\right)=\fbox{$-\frac{26}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{7}}{2}\right)^2=\frac{4}{49} \text{on }\text{the }\text{left }\text{and }-7\times \frac{4}{49}=-\frac{4}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{26}{3}-\frac{4}{7}=-\frac{194}{21}: \\
-6 \left(x^2+\frac{2 x}{3}+\frac{1}{9}\right)-7 \left(y^2-\frac{4 y}{7}+\frac{4}{49}\right)=\fbox{$-\frac{194}{21}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{3}+\frac{1}{9}=\left(x+\frac{1}{3}\right)^2: \\
-6 \fbox{$\left(x+\frac{1}{3}\right)^2$}-7 \left(y^2-\frac{4 y}{7}+\frac{4}{49}\right)=-\frac{194}{21} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{4 y}{7}+\frac{4}{49}=\left(y-\frac{2}{7}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x+\frac{1}{3}\right)^2-7 \fbox{$\left(y-\frac{2}{7}\right)^2$}=-\frac{194}{21} \\
\end{array}
| khanacademy | amps |
Given the equation $7 x^2-4 x-2 y^2+4 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2+4 y+7 x^2-4 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
-2 y^2+4 y+7 x^2-4 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(7 x^2-4 x+\underline{\text{ }}\right)+\left(-2 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(7 x^2-4 x+\underline{\text{ }}\right)=7 \left(x^2-\frac{4 x}{7}+\underline{\text{ }}\right): \\
\fbox{$7 \left(x^2-\frac{4 x}{7}+\underline{\text{ }}\right)$}+\left(-2 y^2+4 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2+4 y+\underline{\text{ }}\right)=-2 \left(y^2-2 y+\underline{\text{ }}\right): \\
7 \left(x^2-\frac{4 x}{7}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2-2 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-4}{7}}{2}\right)^2=\frac{4}{49} \text{on }\text{the }\text{left }\text{and }7\times \frac{4}{49}=\frac{4}{7} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{4}{7}-2=-\frac{10}{7}: \\
7 \left(x^2-\frac{4 x}{7}+\frac{4}{49}\right)-2 \left(y^2-2 y+\underline{\text{ }}\right)=\fbox{$-\frac{10}{7}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-2\times 1=-2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{10}{7}-2=-\frac{24}{7}: \\
7 \left(x^2-\frac{4 x}{7}+\frac{4}{49}\right)-2 \left(y^2-2 y+1\right)=\fbox{$-\frac{24}{7}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{4 x}{7}+\frac{4}{49}=\left(x-\frac{2}{7}\right)^2: \\
7 \fbox{$\left(x-\frac{2}{7}\right)^2$}-2 \left(y^2-2 y+1\right)=-\frac{24}{7} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-2 y+1=(y-1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 7 \left(x-\frac{2}{7}\right)^2-2 \fbox{$(y-1)^2$}=-\frac{24}{7} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-8 x-2 y^2-2 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-2 y-8 x^2-8 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
-2 y^2-2 y-8 x^2-8 x=4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2-8 x+\underline{\text{ }}\right)+\left(-2 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2-8 x+\underline{\text{ }}\right)=-8 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2+x+\underline{\text{ }}\right)$}+\left(-2 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2-2 y+\underline{\text{ }}\right)=-2 \left(y^2+y+\underline{\text{ }}\right): \\
-8 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2+y+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-8}{4}=-2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
4-2=2: \\
-8 \left(x^2+x+\frac{1}{4}\right)-2 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$2$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-2}{4}=-\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
2-\frac{1}{2}=\frac{3}{2}: \\
-8 \left(x^2+x+\frac{1}{4}\right)-2 \left(y^2+y+\frac{1}{4}\right)=\fbox{$\frac{3}{2}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-8 \fbox{$\left(x+\frac{1}{2}\right)^2$}-2 \left(y^2+y+\frac{1}{4}\right)=\frac{3}{2} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x+\frac{1}{2}\right)^2-2 \fbox{$\left(y+\frac{1}{2}\right)^2$}=\frac{3}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2-8 x-5 y^2+y-8=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+y-5 x^2-8 x-8=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }8 \text{to }\text{both }\text{sides}: \\
-5 y^2+y-5 x^2-8 x=8 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2-8 x+\underline{\text{ }}\right)+\left(-5 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2-8 x+\underline{\text{ }}\right)=-5 \left(x^2+\frac{8 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2+\frac{8 x}{5}+\underline{\text{ }}\right)$}+\left(-5 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+8 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right): \\
-5 \left(x^2+\frac{8 x}{5}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+8 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{8}{5}}{2}\right)^2=\frac{16}{25} \text{on }\text{the }\text{left }\text{and }-5\times \frac{16}{25}=-\frac{16}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8-\frac{16}{5}=\frac{24}{5}: \\
-5 \left(x^2+\frac{8 x}{5}+\frac{16}{25}\right)-5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{24}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{-5}{100}=-\frac{1}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{24}{5}-\frac{1}{20}=\frac{19}{4}: \\
-5 \left(x^2+\frac{8 x}{5}+\frac{16}{25}\right)-5 \left(y^2-\frac{y}{5}+\frac{1}{100}\right)=\fbox{$\frac{19}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{8 x}{5}+\frac{16}{25}=\left(x+\frac{4}{5}\right)^2: \\
-5 \fbox{$\left(x+\frac{4}{5}\right)^2$}-5 \left(y^2-\frac{y}{5}+\frac{1}{100}\right)=\frac{19}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{5}+\frac{1}{100}=\left(y-\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x+\frac{4}{5}\right)^2-5 \fbox{$\left(y-\frac{1}{10}\right)^2$}=\frac{19}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2-7 x+2 y^2+3 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2+3 y+9 x^2-7 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
2 y^2+3 y+9 x^2-7 x=7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(9 x^2-7 x+\underline{\text{ }}\right)+\left(2 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2-7 x+\underline{\text{ }}\right)=9 \left(x^2-\frac{7 x}{9}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2-\frac{7 x}{9}+\underline{\text{ }}\right)$}+\left(2 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2+3 y+\underline{\text{ }}\right)=2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right): \\
9 \left(x^2-\frac{7 x}{9}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{9}}{2}\right)^2=\frac{49}{324} \text{on }\text{the }\text{left }\text{and }9\times \frac{49}{324}=\frac{49}{36} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
7+\frac{49}{36}=\frac{301}{36}: \\
9 \left(x^2-\frac{7 x}{9}+\frac{49}{324}\right)+2 \left(y^2+\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{301}{36}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{9}{16}=\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{301}{36}+\frac{9}{8}=\frac{683}{72}: \\
9 \left(x^2-\frac{7 x}{9}+\frac{49}{324}\right)+2 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$\frac{683}{72}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{7 x}{9}+\frac{49}{324}=\left(x-\frac{7}{18}\right)^2: \\
9 \fbox{$\left(x-\frac{7}{18}\right)^2$}+2 \left(y^2+\frac{3 y}{2}+\frac{9}{16}\right)=\frac{683}{72} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{2}+\frac{9}{16}=\left(y+\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x-\frac{7}{18}\right)^2+2 \fbox{$\left(y+\frac{3}{4}\right)^2$}=\frac{683}{72} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2+5 x-5 y^2-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 x^2+5 x+\left(-5 y^2-2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-5 y^2-4 x^2+5 x-2 \text{from }\text{both }\text{sides}: \\
4 x^2-5 x+\left(5 y^2+2\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
5 y^2+4 x^2-5 x=-2 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(4 x^2-5 x+\underline{\text{ }}\right)+5 y^2=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(4 x^2-5 x+\underline{\text{ }}\right)=4 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-\frac{5 x}{4}+\underline{\text{ }}\right)$}+5 y^2=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }4\times \frac{25}{64}=\frac{25}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{25}{16}-2=-\frac{7}{16}: \\
4 \left(x^2-\frac{5 x}{4}+\frac{25}{64}\right)+5 y^2=\fbox{$-\frac{7}{16}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{5 x}{4}+\frac{25}{64}=\left(x-\frac{5}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \fbox{$\left(x-\frac{5}{8}\right)^2$}+5 y^2=-\frac{7}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $-x^2-2 x+8 y^2+10 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2+10 y-x^2-2 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
8 y^2+10 y-x^2-2 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-x^2-2 x+\underline{\text{ }}\right)+\left(8 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-x^2-2 x+\underline{\text{ }}\right)=-\left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$-\left(x^2+2 x+\underline{\text{ }}\right)$}+\left(8 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2+10 y+\underline{\text{ }}\right)=8 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right): \\
-\left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$8 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-1=-1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3-1=2: \\
-\left(x^2+2 x+1\right)+8 \left(y^2+\frac{5 y}{4}+\underline{\text{ }}\right)=\fbox{$2$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }8\times \frac{25}{64}=\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
2+\frac{25}{8}=\frac{41}{8}: \\
-\left(x^2+2 x+1\right)+8 \left(y^2+\frac{5 y}{4}+\frac{25}{64}\right)=\fbox{$\frac{41}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
-\fbox{$(x+1)^2$}+8 \left(y^2+\frac{5 y}{4}+\frac{25}{64}\right)=\frac{41}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{4}+\frac{25}{64}=\left(y+\frac{5}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -(x+1)^2+8 \fbox{$\left(y+\frac{5}{8}\right)^2$}=\frac{41}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-x-6 y^2+y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+y+5 x^2-x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
-6 y^2+y+5 x^2-x=7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2-x+\underline{\text{ }}\right)+\left(-6 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2-x+\underline{\text{ }}\right)=5 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right)$}+\left(-6 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2+y+\underline{\text{ }}\right)=-6 \left(y^2-\frac{y}{6}+\underline{\text{ }}\right): \\
5 \left(x^2-\frac{x}{5}+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{y}{6}+\underline{\text{ }}\right)$}=\underline{\text{ }}+7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{5}{100}=\frac{1}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
7+\frac{1}{20}=\frac{141}{20}: \\
5 \left(x^2-\frac{x}{5}+\frac{1}{100}\right)-6 \left(y^2-\frac{y}{6}+\underline{\text{ }}\right)=\fbox{$\frac{141}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{6}}{2}\right)^2=\frac{1}{144} \text{on }\text{the }\text{left }\text{and }\frac{-6}{144}=-\frac{1}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{141}{20}-\frac{1}{24}=\frac{841}{120}: \\
5 \left(x^2-\frac{x}{5}+\frac{1}{100}\right)-6 \left(y^2-\frac{y}{6}+\frac{1}{144}\right)=\fbox{$\frac{841}{120}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{x}{5}+\frac{1}{100}=\left(x-\frac{1}{10}\right)^2: \\
5 \fbox{$\left(x-\frac{1}{10}\right)^2$}-6 \left(y^2-\frac{y}{6}+\frac{1}{144}\right)=\frac{841}{120} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{6}+\frac{1}{144}=\left(y-\frac{1}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x-\frac{1}{10}\right)^2-6 \fbox{$\left(y-\frac{1}{12}\right)^2$}=\frac{841}{120} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-7 x+8 y^2-4 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-4 y-10 x^2-7 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
8 y^2-4 y-10 x^2-7 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2-7 x+\underline{\text{ }}\right)+\left(8 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2-7 x+\underline{\text{ }}\right)=-10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right)$}+\left(8 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-4 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right): \\
-10 \left(x^2+\frac{7 x}{10}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{10}}{2}\right)^2=\frac{49}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{49}{400}=-\frac{49}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-2-\frac{49}{40}=-\frac{129}{40}: \\
-10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)+8 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{129}{40}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{8}{16}=\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{2}-\frac{129}{40}=-\frac{109}{40}: \\
-10 \left(x^2+\frac{7 x}{10}+\frac{49}{400}\right)+8 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$-\frac{109}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{10}+\frac{49}{400}=\left(x+\frac{7}{20}\right)^2: \\
-10 \fbox{$\left(x+\frac{7}{20}\right)^2$}+8 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=-\frac{109}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{2}+\frac{1}{16}=\left(y-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x+\frac{7}{20}\right)^2+8 \fbox{$\left(y-\frac{1}{4}\right)^2$}=-\frac{109}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2+9 x+10 y^2-9 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2-9 y-8 x^2+9 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
10 y^2-9 y-8 x^2+9 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2+9 x+\underline{\text{ }}\right)+\left(10 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2+9 x+\underline{\text{ }}\right)=-8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right)$}+\left(10 y^2-9 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2-9 y+\underline{\text{ }}\right)=10 \left(y^2-\frac{9 y}{10}+\underline{\text{ }}\right): \\
-8 \left(x^2-\frac{9 x}{8}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2-\frac{9 y}{10}+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{8}}{2}\right)^2=\frac{81}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{81}{256}=-\frac{81}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3-\frac{81}{32}=\frac{15}{32}: \\
-8 \left(x^2-\frac{9 x}{8}+\frac{81}{256}\right)+10 \left(y^2-\frac{9 y}{10}+\underline{\text{ }}\right)=\fbox{$\frac{15}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{10}}{2}\right)^2=\frac{81}{400} \text{on }\text{the }\text{left }\text{and }10\times \frac{81}{400}=\frac{81}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{15}{32}+\frac{81}{40}=\frac{399}{160}: \\
-8 \left(x^2-\frac{9 x}{8}+\frac{81}{256}\right)+10 \left(y^2-\frac{9 y}{10}+\frac{81}{400}\right)=\fbox{$\frac{399}{160}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{9 x}{8}+\frac{81}{256}=\left(x-\frac{9}{16}\right)^2: \\
-8 \fbox{$\left(x-\frac{9}{16}\right)^2$}+10 \left(y^2-\frac{9 y}{10}+\frac{81}{400}\right)=\frac{399}{160} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{9 y}{10}+\frac{81}{400}=\left(y-\frac{9}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x-\frac{9}{16}\right)^2+\text{10 }\fbox{$\left(y-\frac{9}{20}\right)^2$}=\frac{399}{160} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2-10 x-6 y^2+10 y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+10 y-5 x^2-10 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
-6 y^2+10 y-5 x^2-10 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2-10 x+\underline{\text{ }}\right)+\left(-6 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2-10 x+\underline{\text{ }}\right)=-5 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2+2 x+\underline{\text{ }}\right)$}+\left(-6 y^2+10 y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2+10 y+\underline{\text{ }}\right)=-6 \left(y^2-\frac{5 y}{3}+\underline{\text{ }}\right): \\
-5 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{5 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-5\times 1=-5 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10-5=5: \\
-5 \left(x^2+2 x+1\right)-6 \left(y^2-\frac{5 y}{3}+\underline{\text{ }}\right)=\fbox{$5$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{3}}{2}\right)^2=\frac{25}{36} \text{on }\text{the }\text{left }\text{and }-6\times \frac{25}{36}=-\frac{25}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
5-\frac{25}{6}=\frac{5}{6}: \\
-5 \left(x^2+2 x+1\right)-6 \left(y^2-\frac{5 y}{3}+\frac{25}{36}\right)=\fbox{$\frac{5}{6}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
-5 \fbox{$(x+1)^2$}-6 \left(y^2-\frac{5 y}{3}+\frac{25}{36}\right)=\frac{5}{6} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{5 y}{3}+\frac{25}{36}=\left(y-\frac{5}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 (x+1)^2-6 \fbox{$\left(y-\frac{5}{6}\right)^2$}=\frac{5}{6} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2-4 x+y^2-y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2-y-4 x^2-4 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
y^2-y-4 x^2-4 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-4 x^2-4 x+\underline{\text{ }}\right)+\left(y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2-4 x+\underline{\text{ }}\right)=-4 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2+x+\underline{\text{ }}\right)$}+\left(y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-4}{4}=-1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
3-1=2: \\
-4 \left(x^2+x+\frac{1}{4}\right)+\left(y^2-y+\underline{\text{ }}\right)=\fbox{$2$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
2+\frac{1}{4}=\frac{9}{4}: \\
-4 \left(x^2+x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=\fbox{$\frac{9}{4}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-4 \fbox{$\left(x+\frac{1}{2}\right)^2$}+\left(y^2-y+\frac{1}{4}\right)=\frac{9}{4} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x+\frac{1}{2}\right)^2+\fbox{$\left(y-\frac{1}{2}\right)^2$}=\frac{9}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-2 x+3 y^2+3 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+3 y+2 x^2-2 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
3 y^2+3 y+2 x^2-2 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2-2 x+\underline{\text{ }}\right)+\left(3 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-2 x+\underline{\text{ }}\right)=2 \left(x^2-x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-x+\underline{\text{ }}\right)$}+\left(3 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 y^2+3 y+\underline{\text{ }}\right)=3 \left(y^2+y+\underline{\text{ }}\right): \\
2 \left(x^2-x+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+y+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{2}{4}=\frac{1}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{1}{2}-9=-\frac{17}{2}: \\
2 \left(x^2-x+\frac{1}{4}\right)+3 \left(y^2+y+\underline{\text{ }}\right)=\fbox{$-\frac{17}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{3}{4}=\frac{3}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{3}{4}-\frac{17}{2}=-\frac{31}{4}: \\
2 \left(x^2-x+\frac{1}{4}\right)+3 \left(y^2+y+\frac{1}{4}\right)=\fbox{$-\frac{31}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
2 \fbox{$\left(x-\frac{1}{2}\right)^2$}+3 \left(y^2+y+\frac{1}{4}\right)=-\frac{31}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+y+\frac{1}{4}=\left(y+\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x-\frac{1}{2}\right)^2+3 \fbox{$\left(y+\frac{1}{2}\right)^2$}=-\frac{31}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2-4 x+y^2-2 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
y^2-2 y-10 x^2-4 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
y^2-2 y-10 x^2-4 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2-4 x+\underline{\text{ }}\right)+\left(y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2-4 x+\underline{\text{ }}\right)=-10 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right)$}+\left(y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{5}}{2}\right)^2=\frac{1}{25} \text{on }\text{the }\text{left }\text{and }\frac{-10}{25}=-\frac{2}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
-2-\frac{2}{5}=-\frac{12}{5}: \\
-10 \left(x^2+\frac{2 x}{5}+\frac{1}{25}\right)+\left(y^2-2 y+\underline{\text{ }}\right)=\fbox{$-\frac{12}{5}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-2}{2}\right)^2=1 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 8:
\begin{array}{l}
1-\frac{12}{5}=-\frac{7}{5}: \\
-10 \left(x^2+\frac{2 x}{5}+\frac{1}{25}\right)+\left(y^2-2 y+1\right)=\fbox{$-\frac{7}{5}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+\frac{2 x}{5}+\frac{1}{25}=\left(x+\frac{1}{5}\right)^2: \\
-10 \fbox{$\left(x+\frac{1}{5}\right)^2$}+\left(y^2-2 y+1\right)=-\frac{7}{5} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-2 y+1=(y-1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x+\frac{1}{5}\right)^2+\fbox{$(y-1)^2$}=-\frac{7}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2-9 x+10 y^2+8 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2+8 y+5 x^2-9 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
10 y^2+8 y+5 x^2-9 x=4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2-9 x+\underline{\text{ }}\right)+\left(10 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2-9 x+\underline{\text{ }}\right)=5 \left(x^2-\frac{9 x}{5}+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2-\frac{9 x}{5}+\underline{\text{ }}\right)$}+\left(10 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2+8 y+\underline{\text{ }}\right)=10 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right): \\
5 \left(x^2-\frac{9 x}{5}+\underline{\text{ }}\right)+\fbox{$10 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{5}}{2}\right)^2=\frac{81}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{81}{100}=\frac{81}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
4+\frac{81}{20}=\frac{161}{20}: \\
5 \left(x^2-\frac{9 x}{5}+\frac{81}{100}\right)+10 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{161}{20}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }10\times \frac{4}{25}=\frac{8}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{161}{20}+\frac{8}{5}=\frac{193}{20}: \\
5 \left(x^2-\frac{9 x}{5}+\frac{81}{100}\right)+10 \left(y^2+\frac{4 y}{5}+\frac{4}{25}\right)=\fbox{$\frac{193}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{9 x}{5}+\frac{81}{100}=\left(x-\frac{9}{10}\right)^2: \\
5 \fbox{$\left(x-\frac{9}{10}\right)^2$}+10 \left(y^2+\frac{4 y}{5}+\frac{4}{25}\right)=\frac{193}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{4 y}{5}+\frac{4}{25}=\left(y+\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x-\frac{9}{10}\right)^2+\text{10 }\fbox{$\left(y+\frac{2}{5}\right)^2$}=\frac{193}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-6 x+5 y^2+y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+y+x^2-6 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
5 y^2+y+x^2-6 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(x^2-6 x+\underline{\text{ }}\right)+\left(5 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 y^2+y+\underline{\text{ }}\right)=5 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right): \\
\left(x^2-6 x+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-6}{2}\right)^2=9 \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
9-5=4: \\
\left(x^2-6 x+9\right)+5 \left(y^2+\frac{y}{5}+\underline{\text{ }}\right)=\fbox{$4$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{5}{100}=\frac{1}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 8:
\begin{array}{l}
4+\frac{1}{20}=\frac{81}{20}: \\
\left(x^2-6 x+9\right)+5 \left(y^2+\frac{y}{5}+\frac{1}{100}\right)=\fbox{$\frac{81}{20}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-6 x+9=(x-3)^2: \\
\fbox{$(x-3)^2$}+5 \left(y^2+\frac{y}{5}+\frac{1}{100}\right)=\frac{81}{20} \\
\end{array}
Step 10:
\begin{array}{l}
y^2+\frac{y}{5}+\frac{1}{100}=\left(y+\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & (x-3)^2+5 \fbox{$\left(y+\frac{1}{10}\right)^2$}=\frac{81}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2+6 x-y^2+6 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2+6 y-10 x^2+6 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-y^2+6 y-10 x^2+6 x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2+6 x+\underline{\text{ }}\right)+\left(-y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2+6 x+\underline{\text{ }}\right)=-10 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right)$}+\left(-y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2+6 y+\underline{\text{ }}\right)=-\left(y^2-6 y+\underline{\text{ }}\right): \\
-10 \left(x^2-\frac{3 x}{5}+\underline{\text{ }}\right)+\fbox{$-\left(y^2-6 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }-10\times \frac{9}{100}=-\frac{9}{10} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-7-\frac{9}{10}=-\frac{79}{10}: \\
-10 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)-\left(y^2-6 y+\underline{\text{ }}\right)=\fbox{$-\frac{79}{10}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-6}{2}\right)^2=9 \text{on }\text{the }\text{left }\text{and }-9=-9 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{79}{10}-9=-\frac{169}{10}: \\
-10 \left(x^2-\frac{3 x}{5}+\frac{9}{100}\right)-\left(y^2-6 y+9\right)=\fbox{$-\frac{169}{10}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{3 x}{5}+\frac{9}{100}=\left(x-\frac{3}{10}\right)^2: \\
-10 \fbox{$\left(x-\frac{3}{10}\right)^2$}-\left(y^2-6 y+9\right)=-\frac{169}{10} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-6 y+9=(y-3)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x-\frac{3}{10}\right)^2-\fbox{$(y-3)^2$}=-\frac{169}{10} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-7 x+6 y^2-10 y-4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
6 y^2-10 y-3 x^2-7 x-4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }4 \text{to }\text{both }\text{sides}: \\
6 y^2-10 y-3 x^2-7 x=4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2-7 x+\underline{\text{ }}\right)+\left(6 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-7 x+\underline{\text{ }}\right)=-3 \left(x^2+\frac{7 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+\frac{7 x}{3}+\underline{\text{ }}\right)$}+\left(6 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(6 y^2-10 y+\underline{\text{ }}\right)=6 \left(y^2-\frac{5 y}{3}+\underline{\text{ }}\right): \\
-3 \left(x^2+\frac{7 x}{3}+\underline{\text{ }}\right)+\fbox{$6 \left(y^2-\frac{5 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{3}}{2}\right)^2=\frac{49}{36} \text{on }\text{the }\text{left }\text{and }-3\times \frac{49}{36}=-\frac{49}{12} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
4-\frac{49}{12}=-\frac{1}{12}: \\
-3 \left(x^2+\frac{7 x}{3}+\frac{49}{36}\right)+6 \left(y^2-\frac{5 y}{3}+\underline{\text{ }}\right)=\fbox{$-\frac{1}{12}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{3}}{2}\right)^2=\frac{25}{36} \text{on }\text{the }\text{left }\text{and }6\times \frac{25}{36}=\frac{25}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{25}{6}-\frac{1}{12}=\frac{49}{12}: \\
-3 \left(x^2+\frac{7 x}{3}+\frac{49}{36}\right)+6 \left(y^2-\frac{5 y}{3}+\frac{25}{36}\right)=\fbox{$\frac{49}{12}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{3}+\frac{49}{36}=\left(x+\frac{7}{6}\right)^2: \\
-3 \fbox{$\left(x+\frac{7}{6}\right)^2$}+6 \left(y^2-\frac{5 y}{3}+\frac{25}{36}\right)=\frac{49}{12} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{5 y}{3}+\frac{25}{36}=\left(y-\frac{5}{6}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x+\frac{7}{6}\right)^2+6 \fbox{$\left(y-\frac{5}{6}\right)^2$}=\frac{49}{12} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2-8 x-8 y^2+8 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2+8 y+2 x^2-8 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-8 y^2+8 y+2 x^2-8 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2-8 x+\underline{\text{ }}\right)+\left(-8 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2-8 x+\underline{\text{ }}\right)=2 \left(x^2-4 x+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2-4 x+\underline{\text{ }}\right)$}+\left(-8 y^2+8 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2+8 y+\underline{\text{ }}\right)=-8 \left(y^2-y+\underline{\text{ }}\right): \\
2 \left(x^2-4 x+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-4}{2}\right)^2=4 \text{on }\text{the }\text{left }\text{and }2\times 4=8 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
8-5=3: \\
2 \left(x^2-4 x+4\right)-8 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$3$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-8}{4}=-2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
3-2=1: \\
2 \left(x^2-4 x+4\right)-8 \left(y^2-y+\frac{1}{4}\right)=\fbox{$1$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-4 x+4=(x-2)^2: \\
2 \fbox{$(x-2)^2$}-8 \left(y^2-y+\frac{1}{4}\right)=1 \\
\end{array}
Step 11:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 (x-2)^2-8 \fbox{$\left(y-\frac{1}{2}\right)^2$}=1 \\
\end{array}
| khanacademy | amps |
Given the equation $-2 x^2-5 x+4 y^2-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 x^2-5 x+\left(4 y^2-9\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 y^2-2 x^2-5 x-9 \text{from }\text{both }\text{sides}: \\
2 x^2+5 x+\left(9-4 y^2\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-4 y^2+2 x^2+5 x=-9 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(2 x^2+5 x+\underline{\text{ }}\right)-4 y^2=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 x^2+5 x+\underline{\text{ }}\right)=2 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)$}-4 y^2=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{25}{16}=\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{25}{8}-9=-\frac{47}{8}: \\
2 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)-4 y^2=\fbox{$-\frac{47}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+\frac{5 x}{2}+\frac{25}{16}=\left(x+\frac{5}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \fbox{$\left(x+\frac{5}{4}\right)^2$}-4 y^2=-\frac{47}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2+10 x-8 y^2-2 y-6=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2-2 y+4 x^2+10 x-6=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 \text{to }\text{both }\text{sides}: \\
-8 y^2-2 y+4 x^2+10 x=6 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(4 x^2+10 x+\underline{\text{ }}\right)+\left(-8 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2+10 x+\underline{\text{ }}\right)=4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)$}+\left(-8 y^2-2 y+\underline{\text{ }}\right)=\underline{\text{ }}+6 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2-2 y+\underline{\text{ }}\right)=-8 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right): \\
4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+6 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }4\times \frac{25}{16}=\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
6+\frac{25}{4}=\frac{49}{4}: \\
4 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)-8 \left(y^2+\frac{y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{49}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{-8}{64}=-\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{49}{4}-\frac{1}{8}=\frac{97}{8}: \\
4 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)-8 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=\fbox{$\frac{97}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{2}+\frac{25}{16}=\left(x+\frac{5}{4}\right)^2: \\
4 \fbox{$\left(x+\frac{5}{4}\right)^2$}-8 \left(y^2+\frac{y}{4}+\frac{1}{64}\right)=\frac{97}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{4}+\frac{1}{64}=\left(y+\frac{1}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \left(x+\frac{5}{4}\right)^2-8 \fbox{$\left(y+\frac{1}{8}\right)^2$}=\frac{97}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-7 x-10 y^2-8 y+4=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-8 y-8 x^2-7 x+4=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }4 \text{from }\text{both }\text{sides}: \\
-10 y^2-8 y-8 x^2-7 x=-4 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2-7 x+\underline{\text{ }}\right)+\left(-10 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2-7 x+\underline{\text{ }}\right)=-8 \left(x^2+\frac{7 x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2+\frac{7 x}{8}+\underline{\text{ }}\right)$}+\left(-10 y^2-8 y+\underline{\text{ }}\right)=\underline{\text{ }}-4 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2-8 y+\underline{\text{ }}\right)=-10 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right): \\
-8 \left(x^2+\frac{7 x}{8}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-4 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{7}{8}}{2}\right)^2=\frac{49}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{49}{256}=-\frac{49}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-4-\frac{49}{32}=-\frac{177}{32}: \\
-8 \left(x^2+\frac{7 x}{8}+\frac{49}{256}\right)-10 \left(y^2+\frac{4 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{177}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{4}{5}}{2}\right)^2=\frac{4}{25} \text{on }\text{the }\text{left }\text{and }-10\times \frac{4}{25}=-\frac{8}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{177}{32}-\frac{8}{5}=-\frac{1141}{160}: \\
-8 \left(x^2+\frac{7 x}{8}+\frac{49}{256}\right)-10 \left(y^2+\frac{4 y}{5}+\frac{4}{25}\right)=\fbox{$-\frac{1141}{160}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{7 x}{8}+\frac{49}{256}=\left(x+\frac{7}{16}\right)^2: \\
-8 \fbox{$\left(x+\frac{7}{16}\right)^2$}-10 \left(y^2+\frac{4 y}{5}+\frac{4}{25}\right)=-\frac{1141}{160} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{4 y}{5}+\frac{4}{25}=\left(y+\frac{2}{5}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x+\frac{7}{16}\right)^2-\text{10 }\fbox{$\left(y+\frac{2}{5}\right)^2$}=-\frac{1141}{160} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2+3 x-10 y^2-3 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-3 y-6 x^2+3 x=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2+3 x+\underline{\text{ }}\right)+\left(-10 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 3:
\begin{array}{l}
\left(-6 x^2+3 x+\underline{\text{ }}\right)=-6 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)$}+\left(-10 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 y^2-3 y+\underline{\text{ }}\right)=-10 \left(y^2+\frac{3 y}{10}+\underline{\text{ }}\right): \\
-6 \left(x^2-\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+\frac{3 y}{10}+\underline{\text{ }}\right)$}=\underline{\text{ }}+0 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-6}{16}=-\frac{3}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{10}}{2}\right)^2=\frac{9}{400} \text{on }\text{the }\text{left }\text{and }-10\times \frac{9}{400}=-\frac{9}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-\frac{3}{8}-\frac{9}{40}=-\frac{3}{5}: \\
-6 \left(x^2-\frac{x}{2}+\frac{1}{16}\right)-10 \left(y^2+\frac{3 y}{10}+\frac{9}{400}\right)=\fbox{$-\frac{3}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2-\frac{x}{2}+\frac{1}{16}=\left(x-\frac{1}{4}\right)^2: \\
-6 \fbox{$\left(x-\frac{1}{4}\right)^2$}-10 \left(y^2+\frac{3 y}{10}+\frac{9}{400}\right)=-\frac{3}{5} \\
\end{array}
Step 9:
\begin{array}{l}
y^2+\frac{3 y}{10}+\frac{9}{400}=\left(y+\frac{3}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x-\frac{1}{4}\right)^2-\text{10 }\fbox{$\left(y+\frac{3}{20}\right)^2$}=-\frac{3}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2+10 x-5 y^2+9 y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+9 y+6 x^2+10 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
-5 y^2+9 y+6 x^2+10 x=7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(6 x^2+10 x+\underline{\text{ }}\right)+\left(-5 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 x^2+10 x+\underline{\text{ }}\right)=6 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right)$}+\left(-5 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+9 y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{9 y}{5}+\underline{\text{ }}\right): \\
6 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{9 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{3}}{2}\right)^2=\frac{25}{36} \text{on }\text{the }\text{left }\text{and }6\times \frac{25}{36}=\frac{25}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
7+\frac{25}{6}=\frac{67}{6}: \\
6 \left(x^2+\frac{5 x}{3}+\frac{25}{36}\right)-5 \left(y^2-\frac{9 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{67}{6}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{5}}{2}\right)^2=\frac{81}{100} \text{on }\text{the }\text{left }\text{and }-5\times \frac{81}{100}=-\frac{81}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{67}{6}-\frac{81}{20}=\frac{427}{60}: \\
6 \left(x^2+\frac{5 x}{3}+\frac{25}{36}\right)-5 \left(y^2-\frac{9 y}{5}+\frac{81}{100}\right)=\fbox{$\frac{427}{60}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{3}+\frac{25}{36}=\left(x+\frac{5}{6}\right)^2: \\
6 \fbox{$\left(x+\frac{5}{6}\right)^2$}-5 \left(y^2-\frac{9 y}{5}+\frac{81}{100}\right)=\frac{427}{60} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{9 y}{5}+\frac{81}{100}=\left(y-\frac{9}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x+\frac{5}{6}\right)^2-5 \fbox{$\left(y-\frac{9}{10}\right)^2$}=\frac{427}{60} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2-9 x-3 y^2+3 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2+3 y-7 x^2-9 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-3 y^2+3 y-7 x^2-9 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-7 x^2-9 x+\underline{\text{ }}\right)+\left(-3 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2-9 x+\underline{\text{ }}\right)=-7 \left(x^2+\frac{9 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2+\frac{9 x}{7}+\underline{\text{ }}\right)$}+\left(-3 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2+3 y+\underline{\text{ }}\right)=-3 \left(y^2-y+\underline{\text{ }}\right): \\
-7 \left(x^2+\frac{9 x}{7}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2-y+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{7}}{2}\right)^2=\frac{81}{196} \text{on }\text{the }\text{left }\text{and }-7\times \frac{81}{196}=-\frac{81}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-1-\frac{81}{28}=-\frac{109}{28}: \\
-7 \left(x^2+\frac{9 x}{7}+\frac{81}{196}\right)-3 \left(y^2-y+\underline{\text{ }}\right)=\fbox{$-\frac{109}{28}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-3}{4}=-\frac{3}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{109}{28}-\frac{3}{4}=-\frac{65}{14}: \\
-7 \left(x^2+\frac{9 x}{7}+\frac{81}{196}\right)-3 \left(y^2-y+\frac{1}{4}\right)=\fbox{$-\frac{65}{14}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{9 x}{7}+\frac{81}{196}=\left(x+\frac{9}{14}\right)^2: \\
-7 \fbox{$\left(x+\frac{9}{14}\right)^2$}-3 \left(y^2-y+\frac{1}{4}\right)=-\frac{65}{14} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-y+\frac{1}{4}=\left(y-\frac{1}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x+\frac{9}{14}\right)^2-3 \fbox{$\left(y-\frac{1}{2}\right)^2$}=-\frac{65}{14} \\
\end{array}
| khanacademy | amps |
Given the equation $10 x^2-10 x-2 y^2-5 y+7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-5 y+10 x^2-10 x+7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }7 \text{from }\text{both }\text{sides}: \\
-2 y^2-5 y+10 x^2-10 x=-7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(10 x^2-10 x+\underline{\text{ }}\right)+\left(-2 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(10 x^2-10 x+\underline{\text{ }}\right)=10 \left(x^2-x+\underline{\text{ }}\right): \\
\fbox{$10 \left(x^2-x+\underline{\text{ }}\right)$}+\left(-2 y^2-5 y+\underline{\text{ }}\right)=\underline{\text{ }}-7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2-5 y+\underline{\text{ }}\right)=-2 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right): \\
10 \left(x^2-x+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}-7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{10}{4}=\frac{5}{2} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{5}{2}-7=-\frac{9}{2}: \\
10 \left(x^2-x+\frac{1}{4}\right)-2 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$-\frac{9}{2}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }-2\times \frac{25}{16}=-\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{9}{2}-\frac{25}{8}=-\frac{61}{8}: \\
10 \left(x^2-x+\frac{1}{4}\right)-2 \left(y^2+\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{61}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2: \\
\text{10 }\fbox{$\left(x-\frac{1}{2}\right)^2$}-2 \left(y^2+\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{61}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{2}+\frac{25}{16}=\left(y+\frac{5}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 10 \left(x-\frac{1}{2}\right)^2-2 \fbox{$\left(y+\frac{5}{4}\right)^2$}=-\frac{61}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-10 x^2+4 x-10 y^2-6 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2-6 y-10 x^2+4 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-10 y^2-6 y-10 x^2+4 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-10 x^2+4 x+\underline{\text{ }}\right)+\left(-10 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 x^2+4 x+\underline{\text{ }}\right)=-10 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-10 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right)$}+\left(-10 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-10 y^2-6 y+\underline{\text{ }}\right)=-10 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right): \\
-10 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{5}}{2}\right)^2=\frac{1}{25} \text{on }\text{the }\text{left }\text{and }\frac{-10}{25}=-\frac{2}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-9-\frac{2}{5}=-\frac{47}{5}: \\
-10 \left(x^2-\frac{2 x}{5}+\frac{1}{25}\right)-10 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{47}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }-10\times \frac{9}{100}=-\frac{9}{10} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{47}{5}-\frac{9}{10}=-\frac{103}{10}: \\
-10 \left(x^2-\frac{2 x}{5}+\frac{1}{25}\right)-10 \left(y^2+\frac{3 y}{5}+\frac{9}{100}\right)=\fbox{$-\frac{103}{10}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{2 x}{5}+\frac{1}{25}=\left(x-\frac{1}{5}\right)^2: \\
-10 \fbox{$\left(x-\frac{1}{5}\right)^2$}-10 \left(y^2+\frac{3 y}{5}+\frac{9}{100}\right)=-\frac{103}{10} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{5}+\frac{9}{100}=\left(y+\frac{3}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -10 \left(x-\frac{1}{5}\right)^2-\text{10 }\fbox{$\left(y+\frac{3}{10}\right)^2$}=-\frac{103}{10} \\
\end{array}
| khanacademy | amps |
Given the equation $4 x^2-6 x-10 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
4 x^2-6 x+(1-10 y)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1-10 y \text{from }\text{both }\text{sides}: \\
4 x^2-6 x=10 y-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(4 x^2-6 x+\underline{\text{ }}\right)=(10 y-1)+\underline{\text{ }} \\
\end{array}
Step 4:
\begin{array}{l}
\left(4 x^2-6 x+\underline{\text{ }}\right)=4 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right): \\
\fbox{$4 \left(x^2-\frac{3 x}{2}+\underline{\text{ }}\right)$}=(10 y-1)+\underline{\text{ }} \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }4\times \frac{9}{16}=\frac{9}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
(10 y-1)+\frac{9}{4}=10 y+\frac{5}{4}: \\
4 \left(x^2-\frac{3 x}{2}+\frac{9}{16}\right)=\fbox{$10 y+\frac{5}{4}$} \\
\end{array}
Step 7:
\begin{array}{l}
x^2-\frac{3 x}{2}+\frac{9}{16}=\left(x-\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 4 \fbox{$\left(x-\frac{3}{4}\right)^2$}=10 y+\frac{5}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-3 x+3 y^2+2 y-2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
3 y^2+2 y-6 x^2-3 x-2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }2 \text{to }\text{both }\text{sides}: \\
3 y^2+2 y-6 x^2-3 x=2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2-3 x+\underline{\text{ }}\right)+\left(3 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2-3 x+\underline{\text{ }}\right)=-6 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)$}+\left(3 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 y^2+2 y+\underline{\text{ }}\right)=3 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right): \\
-6 \left(x^2+\frac{x}{2}+\underline{\text{ }}\right)+\fbox{$3 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right)$}=\underline{\text{ }}+2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-6}{16}=-\frac{3}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
2-\frac{3}{8}=\frac{13}{8}: \\
-6 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+3 \left(y^2+\frac{2 y}{3}+\underline{\text{ }}\right)=\fbox{$\frac{13}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{3}{9}=\frac{1}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{13}{8}+\frac{1}{3}=\frac{47}{24}: \\
-6 \left(x^2+\frac{x}{2}+\frac{1}{16}\right)+3 \left(y^2+\frac{2 y}{3}+\frac{1}{9}\right)=\fbox{$\frac{47}{24}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{2}+\frac{1}{16}=\left(x+\frac{1}{4}\right)^2: \\
-6 \fbox{$\left(x+\frac{1}{4}\right)^2$}+3 \left(y^2+\frac{2 y}{3}+\frac{1}{9}\right)=\frac{47}{24} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{2 y}{3}+\frac{1}{9}=\left(y+\frac{1}{3}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x+\frac{1}{4}\right)^2+3 \fbox{$\left(y+\frac{1}{3}\right)^2$}=\frac{47}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2-10 x-8 y^2+2 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2+2 y-5 x^2-10 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-8 y^2+2 y-5 x^2-10 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2-10 x+\underline{\text{ }}\right)+\left(-8 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2-10 x+\underline{\text{ }}\right)=-5 \left(x^2+2 x+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2+2 x+\underline{\text{ }}\right)$}+\left(-8 y^2+2 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2+2 y+\underline{\text{ }}\right)=-8 \left(y^2-\frac{y}{4}+\underline{\text{ }}\right): \\
-5 \left(x^2+2 x+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2-\frac{y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-5\times 1=-5 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9-5=4: \\
-5 \left(x^2+2 x+1\right)-8 \left(y^2-\frac{y}{4}+\underline{\text{ }}\right)=\fbox{$4$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{4}}{2}\right)^2=\frac{1}{64} \text{on }\text{the }\text{left }\text{and }\frac{-8}{64}=-\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
4-\frac{1}{8}=\frac{31}{8}: \\
-5 \left(x^2+2 x+1\right)-8 \left(y^2-\frac{y}{4}+\frac{1}{64}\right)=\fbox{$\frac{31}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+2 x+1=(x+1)^2: \\
-5 \fbox{$(x+1)^2$}-8 \left(y^2-\frac{y}{4}+\frac{1}{64}\right)=\frac{31}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{4}+\frac{1}{64}=\left(y-\frac{1}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 (x+1)^2-8 \fbox{$\left(y-\frac{1}{8}\right)^2$}=\frac{31}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2-2 x-2 y^2-4 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-2 y^2-4 y-5 x^2-2 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-2 y^2-4 y-5 x^2-2 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2-2 x+\underline{\text{ }}\right)+\left(-2 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2-2 x+\underline{\text{ }}\right)=-5 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right)$}+\left(-2 y^2-4 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-2 y^2-4 y+\underline{\text{ }}\right)=-2 \left(y^2+2 y+\underline{\text{ }}\right): \\
-5 \left(x^2+\frac{2 x}{5}+\underline{\text{ }}\right)+\fbox{$-2 \left(y^2+2 y+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{2}{5}}{2}\right)^2=\frac{1}{25} \text{on }\text{the }\text{left }\text{and }\frac{-5}{25}=-\frac{1}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9-\frac{1}{5}=\frac{44}{5}: \\
-5 \left(x^2+\frac{2 x}{5}+\frac{1}{25}\right)-2 \left(y^2+2 y+\underline{\text{ }}\right)=\fbox{$\frac{44}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-2\times 1=-2 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{44}{5}-2=\frac{34}{5}: \\
-5 \left(x^2+\frac{2 x}{5}+\frac{1}{25}\right)-2 \left(y^2+2 y+1\right)=\fbox{$\frac{34}{5}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{2 x}{5}+\frac{1}{25}=\left(x+\frac{1}{5}\right)^2: \\
-5 \fbox{$\left(x+\frac{1}{5}\right)^2$}-2 \left(y^2+2 y+1\right)=\frac{34}{5} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x+\frac{1}{5}\right)^2-2 \fbox{$(y+1)^2$}=\frac{34}{5} \\
\end{array}
| khanacademy | amps |
Given the equation $2 x^2+5 x-7 y^2+y-7=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-7 y^2+y+2 x^2+5 x-7=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }7 \text{to }\text{both }\text{sides}: \\
-7 y^2+y+2 x^2+5 x=7 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(2 x^2+5 x+\underline{\text{ }}\right)+\left(-7 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 4:
\begin{array}{l}
\left(2 x^2+5 x+\underline{\text{ }}\right)=2 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$2 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)$}+\left(-7 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+7 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-7 y^2+y+\underline{\text{ }}\right)=-7 \left(y^2-\frac{y}{7}+\underline{\text{ }}\right): \\
2 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$-7 \left(y^2-\frac{y}{7}+\underline{\text{ }}\right)$}=\underline{\text{ }}+7 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }2\times \frac{25}{16}=\frac{25}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
7+\frac{25}{8}=\frac{81}{8}: \\
2 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)-7 \left(y^2-\frac{y}{7}+\underline{\text{ }}\right)=\fbox{$\frac{81}{8}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{7}}{2}\right)^2=\frac{1}{196} \text{on }\text{the }\text{left }\text{and }\frac{-7}{196}=-\frac{1}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{81}{8}-\frac{1}{28}=\frac{565}{56}: \\
2 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)-7 \left(y^2-\frac{y}{7}+\frac{1}{196}\right)=\fbox{$\frac{565}{56}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{2}+\frac{25}{16}=\left(x+\frac{5}{4}\right)^2: \\
2 \fbox{$\left(x+\frac{5}{4}\right)^2$}-7 \left(y^2-\frac{y}{7}+\frac{1}{196}\right)=\frac{565}{56} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{7}+\frac{1}{196}=\left(y-\frac{1}{14}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 2 \left(x+\frac{5}{4}\right)^2-7 \fbox{$\left(y-\frac{1}{14}\right)^2$}=\frac{565}{56} \\
\end{array}
| khanacademy | amps |
Given the equation $-9 x^2-6 y^2-5 y=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2-5 y-9 x^2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }6 y^2+5 y+9 x^2 \text{to }\text{both }\text{sides}: \\
6 y^2+5 y+9 x^2=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(6 y^2+5 y+\underline{\text{ }}\right)+9 x^2=\underline{\text{ }}+0 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 y^2+5 y+\underline{\text{ }}\right)=6 \left(y^2+\frac{5 y}{6}+\underline{\text{ }}\right): \\
\fbox{$6 \left(y^2+\frac{5 y}{6}+\underline{\text{ }}\right)$}+9 x^2=\underline{\text{ }}+0 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{6}}{2}\right)^2=\frac{25}{144} \text{on }\text{the }\text{left }\text{and }6\times \frac{25}{144}=\frac{25}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 6:
\begin{array}{l}
y^2+\frac{5 y}{6}+\frac{25}{144}=\left(y+\frac{5}{12}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \fbox{$\left(y+\frac{5}{12}\right)^2$}+9 x^2=\frac{25}{24} \\
\end{array}
| khanacademy | amps |
Given the equation $-8 x^2-9 x+8 y^2+y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2+y-8 x^2-9 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
8 y^2+y-8 x^2-9 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-8 x^2-9 x+\underline{\text{ }}\right)+\left(8 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 x^2-9 x+\underline{\text{ }}\right)=-8 \left(x^2+\frac{9 x}{8}+\underline{\text{ }}\right): \\
\fbox{$-8 \left(x^2+\frac{9 x}{8}+\underline{\text{ }}\right)$}+\left(8 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2+y+\underline{\text{ }}\right)=8 \left(y^2+\frac{y}{8}+\underline{\text{ }}\right): \\
-8 \left(x^2+\frac{9 x}{8}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2+\frac{y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{9}{8}}{2}\right)^2=\frac{81}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{81}{256}=-\frac{81}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-2-\frac{81}{32}=-\frac{145}{32}: \\
-8 \left(x^2+\frac{9 x}{8}+\frac{81}{256}\right)+8 \left(y^2+\frac{y}{8}+\underline{\text{ }}\right)=\fbox{$-\frac{145}{32}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{8}}{2}\right)^2=\frac{1}{256} \text{on }\text{the }\text{left }\text{and }\frac{8}{256}=\frac{1}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{1}{32}-\frac{145}{32}=-\frac{9}{2}: \\
-8 \left(x^2+\frac{9 x}{8}+\frac{81}{256}\right)+8 \left(y^2+\frac{y}{8}+\frac{1}{256}\right)=\fbox{$-\frac{9}{2}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{9 x}{8}+\frac{81}{256}=\left(x+\frac{9}{16}\right)^2: \\
-8 \fbox{$\left(x+\frac{9}{16}\right)^2$}+8 \left(y^2+\frac{y}{8}+\frac{1}{256}\right)=-\frac{9}{2} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{8}+\frac{1}{256}=\left(y+\frac{1}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -8 \left(x+\frac{9}{16}\right)^2+8 \fbox{$\left(y+\frac{1}{16}\right)^2$}=-\frac{9}{2} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2+8 x+5 y^2-7 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2-7 y-5 x^2+8 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
5 y^2-7 y-5 x^2+8 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2+8 x+\underline{\text{ }}\right)+\left(5 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2+8 x+\underline{\text{ }}\right)=-5 \left(x^2-\frac{8 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2-\frac{8 x}{5}+\underline{\text{ }}\right)$}+\left(5 y^2-7 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2-7 y+\underline{\text{ }}\right)=5 \left(y^2-\frac{7 y}{5}+\underline{\text{ }}\right): \\
-5 \left(x^2-\frac{8 x}{5}+\underline{\text{ }}\right)+\fbox{$5 \left(y^2-\frac{7 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-8}{5}}{2}\right)^2=\frac{16}{25} \text{on }\text{the }\text{left }\text{and }-5\times \frac{16}{25}=-\frac{16}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9-\frac{16}{5}=\frac{29}{5}: \\
-5 \left(x^2-\frac{8 x}{5}+\frac{16}{25}\right)+5 \left(y^2-\frac{7 y}{5}+\underline{\text{ }}\right)=\fbox{$\frac{29}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{5}}{2}\right)^2=\frac{49}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{49}{100}=\frac{49}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{29}{5}+\frac{49}{20}=\frac{33}{4}: \\
-5 \left(x^2-\frac{8 x}{5}+\frac{16}{25}\right)+5 \left(y^2-\frac{7 y}{5}+\frac{49}{100}\right)=\fbox{$\frac{33}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{8 x}{5}+\frac{16}{25}=\left(x-\frac{4}{5}\right)^2: \\
-5 \fbox{$\left(x-\frac{4}{5}\right)^2$}+5 \left(y^2-\frac{7 y}{5}+\frac{49}{100}\right)=\frac{33}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{7 y}{5}+\frac{49}{100}=\left(y-\frac{7}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x-\frac{4}{5}\right)^2+5 \fbox{$\left(y-\frac{7}{10}\right)^2$}=\frac{33}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2-4 x+10 y^2+y-5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
10 y^2+y-4 x^2-4 x-5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }5 \text{to }\text{both }\text{sides}: \\
10 y^2+y-4 x^2-4 x=5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-4 x^2-4 x+\underline{\text{ }}\right)+\left(10 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2-4 x+\underline{\text{ }}\right)=-4 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2+x+\underline{\text{ }}\right)$}+\left(10 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}+5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(10 y^2+y+\underline{\text{ }}\right)=10 \left(y^2+\frac{y}{10}+\underline{\text{ }}\right): \\
-4 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$10 \left(y^2+\frac{y}{10}+\underline{\text{ }}\right)$}=\underline{\text{ }}+5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-4}{4}=-1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
5-1=4: \\
-4 \left(x^2+x+\frac{1}{4}\right)+10 \left(y^2+\frac{y}{10}+\underline{\text{ }}\right)=\fbox{$4$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }\text{10 }\text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{10}}{2}\right)^2=\frac{1}{400} \text{on }\text{the }\text{left }\text{and }\frac{10}{400}=\frac{1}{40} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
4+\frac{1}{40}=\frac{161}{40}: \\
-4 \left(x^2+x+\frac{1}{4}\right)+10 \left(y^2+\frac{y}{10}+\frac{1}{400}\right)=\fbox{$\frac{161}{40}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-4 \fbox{$\left(x+\frac{1}{2}\right)^2$}+10 \left(y^2+\frac{y}{10}+\frac{1}{400}\right)=\frac{161}{40} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{y}{10}+\frac{1}{400}=\left(y+\frac{1}{20}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x+\frac{1}{2}\right)^2+\text{10 }\fbox{$\left(y+\frac{1}{20}\right)^2$}=\frac{161}{40} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-5 x+8 y^2-3 y+2=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
8 y^2-3 y-3 x^2-5 x+2=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }2 \text{from }\text{both }\text{sides}: \\
8 y^2-3 y-3 x^2-5 x=-2 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2-5 x+\underline{\text{ }}\right)+\left(8 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2-5 x+\underline{\text{ }}\right)=-3 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right)$}+\left(8 y^2-3 y+\underline{\text{ }}\right)=\underline{\text{ }}-2 \\
\end{array}
Step 5:
\begin{array}{l}
\left(8 y^2-3 y+\underline{\text{ }}\right)=8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right): \\
-3 \left(x^2+\frac{5 x}{3}+\underline{\text{ }}\right)+\fbox{$8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}-2 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{3}}{2}\right)^2=\frac{25}{36} \text{on }\text{the }\text{left }\text{and }-3\times \frac{25}{36}=-\frac{25}{12} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-2-\frac{25}{12}=-\frac{49}{12}: \\
-3 \left(x^2+\frac{5 x}{3}+\frac{25}{36}\right)+8 \left(y^2-\frac{3 y}{8}+\underline{\text{ }}\right)=\fbox{$-\frac{49}{12}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{8}}{2}\right)^2=\frac{9}{256} \text{on }\text{the }\text{left }\text{and }8\times \frac{9}{256}=\frac{9}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{9}{32}-\frac{49}{12}=-\frac{365}{96}: \\
-3 \left(x^2+\frac{5 x}{3}+\frac{25}{36}\right)+8 \left(y^2-\frac{3 y}{8}+\frac{9}{256}\right)=\fbox{$-\frac{365}{96}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{3}+\frac{25}{36}=\left(x+\frac{5}{6}\right)^2: \\
-3 \fbox{$\left(x+\frac{5}{6}\right)^2$}+8 \left(y^2-\frac{3 y}{8}+\frac{9}{256}\right)=-\frac{365}{96} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{8}+\frac{9}{256}=\left(y-\frac{3}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x+\frac{5}{6}\right)^2+8 \fbox{$\left(y-\frac{3}{16}\right)^2$}=-\frac{365}{96} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2+10 x-8 y^2+9 y+5=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2+9 y-3 x^2+10 x+5=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }5 \text{from }\text{both }\text{sides}: \\
-8 y^2+9 y-3 x^2+10 x=-5 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-3 x^2+10 x+\underline{\text{ }}\right)+\left(-8 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-3 x^2+10 x+\underline{\text{ }}\right)=-3 \left(x^2-\frac{10 x}{3}+\underline{\text{ }}\right): \\
\fbox{$-3 \left(x^2-\frac{10 x}{3}+\underline{\text{ }}\right)$}+\left(-8 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}-5 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-8 y^2+9 y+\underline{\text{ }}\right)=-8 \left(y^2-\frac{9 y}{8}+\underline{\text{ }}\right): \\
-3 \left(x^2-\frac{10 x}{3}+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2-\frac{9 y}{8}+\underline{\text{ }}\right)$}=\underline{\text{ }}-5 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-10}{3}}{2}\right)^2=\frac{25}{9} \text{on }\text{the }\text{left }\text{and }-3\times \frac{25}{9}=-\frac{25}{3} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-5-\frac{25}{3}=-\frac{40}{3}: \\
-3 \left(x^2-\frac{10 x}{3}+\frac{25}{9}\right)-8 \left(y^2-\frac{9 y}{8}+\underline{\text{ }}\right)=\fbox{$-\frac{40}{3}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-9}{8}}{2}\right)^2=\frac{81}{256} \text{on }\text{the }\text{left }\text{and }-8\times \frac{81}{256}=-\frac{81}{32} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{40}{3}-\frac{81}{32}=-\frac{1523}{96}: \\
-3 \left(x^2-\frac{10 x}{3}+\frac{25}{9}\right)-8 \left(y^2-\frac{9 y}{8}+\frac{81}{256}\right)=\fbox{$-\frac{1523}{96}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{10 x}{3}+\frac{25}{9}=\left(x-\frac{5}{3}\right)^2: \\
-3 \fbox{$\left(x-\frac{5}{3}\right)^2$}-8 \left(y^2-\frac{9 y}{8}+\frac{81}{256}\right)=-\frac{1523}{96} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{9 y}{8}+\frac{81}{256}=\left(y-\frac{9}{16}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -3 \left(x-\frac{5}{3}\right)^2-8 \fbox{$\left(y-\frac{9}{16}\right)^2$}=-\frac{1523}{96} \\
\end{array}
| khanacademy | amps |
Given the equation $-7 x^2+5 x+2 y^2-y-10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
2 y^2-y-7 x^2+5 x-10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }\text{10 }\text{to }\text{both }\text{sides}: \\
2 y^2-y-7 x^2+5 x=10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-7 x^2+5 x+\underline{\text{ }}\right)+\left(2 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-7 x^2+5 x+\underline{\text{ }}\right)=-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right): \\
\fbox{$-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right)$}+\left(2 y^2-y+\underline{\text{ }}\right)=\underline{\text{ }}+10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(2 y^2-y+\underline{\text{ }}\right)=2 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right): \\
-7 \left(x^2-\frac{5 x}{7}+\underline{\text{ }}\right)+\fbox{$2 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-7 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{7}}{2}\right)^2=\frac{25}{196} \text{on }\text{the }\text{left }\text{and }-7\times \frac{25}{196}=-\frac{25}{28} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
10-\frac{25}{28}=\frac{255}{28}: \\
-7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)+2 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{255}{28}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }2 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{2}{16}=\frac{1}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{255}{28}+\frac{1}{8}=\frac{517}{56}: \\
-7 \left(x^2-\frac{5 x}{7}+\frac{25}{196}\right)+2 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{517}{56}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{5 x}{7}+\frac{25}{196}=\left(x-\frac{5}{14}\right)^2: \\
-7 \fbox{$\left(x-\frac{5}{14}\right)^2$}+2 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\frac{517}{56} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{2}+\frac{1}{16}=\left(y-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -7 \left(x-\frac{5}{14}\right)^2+2 \fbox{$\left(y-\frac{1}{4}\right)^2$}=\frac{517}{56} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2-7 x-8 y^2-6 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-8 y^2-6 y+x^2-7 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
-8 y^2-6 y+x^2-7 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(x^2-7 x+\underline{\text{ }}\right)+\left(-8 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-8 y^2-6 y+\underline{\text{ }}\right)=-8 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right): \\
\left(x^2-7 x+\underline{\text{ }}\right)+\fbox{$-8 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{-7}{2}\right)^2=\frac{49}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
3+\frac{49}{4}=\frac{61}{4}: \\
\left(x^2-7 x+\frac{49}{4}\right)-8 \left(y^2+\frac{3 y}{4}+\underline{\text{ }}\right)=\fbox{$\frac{61}{4}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-8 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{4}}{2}\right)^2=\frac{9}{64} \text{on }\text{the }\text{left }\text{and }-8\times \frac{9}{64}=-\frac{9}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{61}{4}-\frac{9}{8}=\frac{113}{8}: \\
\left(x^2-7 x+\frac{49}{4}\right)-8 \left(y^2+\frac{3 y}{4}+\frac{9}{64}\right)=\fbox{$\frac{113}{8}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2-7 x+\frac{49}{4}=\left(x-\frac{7}{2}\right)^2: \\
\fbox{$\left(x-\frac{7}{2}\right)^2$}-8 \left(y^2+\frac{3 y}{4}+\frac{9}{64}\right)=\frac{113}{8} \\
\end{array}
Step 10:
\begin{array}{l}
y^2+\frac{3 y}{4}+\frac{9}{64}=\left(y+\frac{3}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \left(x-\frac{7}{2}\right)^2-8 \fbox{$\left(y+\frac{3}{8}\right)^2$}=\frac{113}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2+2 x-4 y^2-10 y-1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2-10 y-5 x^2+2 x-1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
-4 y^2-10 y-5 x^2+2 x=1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2+2 x+\underline{\text{ }}\right)+\left(-4 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2+2 x+\underline{\text{ }}\right)=-5 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right)$}+\left(-4 y^2-10 y+\underline{\text{ }}\right)=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2-10 y+\underline{\text{ }}\right)=-4 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right): \\
-5 \left(x^2-\frac{2 x}{5}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{5}}{2}\right)^2=\frac{1}{25} \text{on }\text{the }\text{left }\text{and }\frac{-5}{25}=-\frac{1}{5} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
1-\frac{1}{5}=\frac{4}{5}: \\
-5 \left(x^2-\frac{2 x}{5}+\frac{1}{25}\right)-4 \left(y^2+\frac{5 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{4}{5}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }-4\times \frac{25}{16}=-\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{4}{5}-\frac{25}{4}=-\frac{109}{20}: \\
-5 \left(x^2-\frac{2 x}{5}+\frac{1}{25}\right)-4 \left(y^2+\frac{5 y}{2}+\frac{25}{16}\right)=\fbox{$-\frac{109}{20}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{2 x}{5}+\frac{1}{25}=\left(x-\frac{1}{5}\right)^2: \\
-5 \fbox{$\left(x-\frac{1}{5}\right)^2$}-4 \left(y^2+\frac{5 y}{2}+\frac{25}{16}\right)=-\frac{109}{20} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{5 y}{2}+\frac{25}{16}=\left(y+\frac{5}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x-\frac{1}{5}\right)^2-4 \fbox{$\left(y+\frac{5}{4}\right)^2$}=-\frac{109}{20} \\
\end{array}
| khanacademy | amps |
Given the equation $-6 x^2-2 x-3 y^2-6 y+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 y^2-6 y-6 x^2-2 x+1=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }1 \text{from }\text{both }\text{sides}: \\
-3 y^2-6 y-6 x^2-2 x=-1 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-6 x^2-2 x+\underline{\text{ }}\right)+\left(-3 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-6 x^2-2 x+\underline{\text{ }}\right)=-6 \left(x^2+\frac{x}{3}+\underline{\text{ }}\right): \\
\fbox{$-6 \left(x^2+\frac{x}{3}+\underline{\text{ }}\right)$}+\left(-3 y^2-6 y+\underline{\text{ }}\right)=\underline{\text{ }}-1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-3 y^2-6 y+\underline{\text{ }}\right)=-3 \left(y^2+2 y+\underline{\text{ }}\right): \\
-6 \left(x^2+\frac{x}{3}+\underline{\text{ }}\right)+\fbox{$-3 \left(y^2+2 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{1}{3}}{2}\right)^2=\frac{1}{36} \text{on }\text{the }\text{left }\text{and }\frac{-6}{36}=-\frac{1}{6} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-1-\frac{1}{6}=-\frac{7}{6}: \\
-6 \left(x^2+\frac{x}{3}+\frac{1}{36}\right)-3 \left(y^2+2 y+\underline{\text{ }}\right)=\fbox{$-\frac{7}{6}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{2}{2}\right)^2=1 \text{on }\text{the }\text{left }\text{and }-3\times 1=-3 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{7}{6}-3=-\frac{25}{6}: \\
-6 \left(x^2+\frac{x}{3}+\frac{1}{36}\right)-3 \left(y^2+2 y+1\right)=\fbox{$-\frac{25}{6}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{x}{3}+\frac{1}{36}=\left(x+\frac{1}{6}\right)^2: \\
-6 \fbox{$\left(x+\frac{1}{6}\right)^2$}-3 \left(y^2+2 y+1\right)=-\frac{25}{6} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+2 y+1=(y+1)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -6 \left(x+\frac{1}{6}\right)^2-3 \fbox{$(y+1)^2$}=-\frac{25}{6} \\
\end{array}
| khanacademy | amps |
Given the equation $9 x^2-6 x-4 y^2+5 y-9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-4 y^2+5 y+9 x^2-6 x-9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }9 \text{to }\text{both }\text{sides}: \\
-4 y^2+5 y+9 x^2-6 x=9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(9 x^2-6 x+\underline{\text{ }}\right)+\left(-4 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(9 x^2-6 x+\underline{\text{ }}\right)=9 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right): \\
\fbox{$9 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)$}+\left(-4 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-4 y^2+5 y+\underline{\text{ }}\right)=-4 \left(y^2-\frac{5 y}{4}+\underline{\text{ }}\right): \\
9 \left(x^2-\frac{2 x}{3}+\underline{\text{ }}\right)+\fbox{$-4 \left(y^2-\frac{5 y}{4}+\underline{\text{ }}\right)$}=\underline{\text{ }}+9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }9 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-2}{3}}{2}\right)^2=\frac{1}{9} \text{on }\text{the }\text{left }\text{and }\frac{9}{9}=1 \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
9+1=10: \\
9 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)-4 \left(y^2-\frac{5 y}{4}+\underline{\text{ }}\right)=\fbox{$10$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-5}{4}}{2}\right)^2=\frac{25}{64} \text{on }\text{the }\text{left }\text{and }-4\times \frac{25}{64}=-\frac{25}{16} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
10-\frac{25}{16}=\frac{135}{16}: \\
9 \left(x^2-\frac{2 x}{3}+\frac{1}{9}\right)-4 \left(y^2-\frac{5 y}{4}+\frac{25}{64}\right)=\fbox{$\frac{135}{16}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{2 x}{3}+\frac{1}{9}=\left(x-\frac{1}{3}\right)^2: \\
9 \fbox{$\left(x-\frac{1}{3}\right)^2$}-4 \left(y^2-\frac{5 y}{4}+\frac{25}{64}\right)=\frac{135}{16} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{5 y}{4}+\frac{25}{64}=\left(y-\frac{5}{8}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 9 \left(x-\frac{1}{3}\right)^2-4 \fbox{$\left(y-\frac{5}{8}\right)^2$}=\frac{135}{16} \\
\end{array}
| khanacademy | amps |
Given the equation $x^2+9 x-10 y^2+5 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-10 y^2+5 y+x^2+9 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
-10 y^2+5 y+x^2+9 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(x^2+9 x+\underline{\text{ }}\right)+\left(-10 y^2+5 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-10 y^2+5 y+\underline{\text{ }}\right)=-10 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right): \\
\left(x^2+9 x+\underline{\text{ }}\right)+\fbox{$-10 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it, }\text{then }\text{add }\text{it }\text{to }\text{both }\text{sides.} \\
\text{Add }\left(\frac{9}{2}\right)^2=\frac{81}{4} \text{to }\text{both }\text{sides}: \\
\end{array}
Step 6:
\begin{array}{l}
3+\frac{81}{4}=\frac{93}{4}: \\
\left(x^2+9 x+\frac{81}{4}\right)-10 \left(y^2-\frac{y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{93}{4}$} \\
\end{array}
Step 7:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-10 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{2}}{2}\right)^2=\frac{1}{16} \text{on }\text{the }\text{left }\text{and }\frac{-10}{16}=-\frac{5}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 8:
\begin{array}{l}
\frac{93}{4}-\frac{5}{8}=\frac{181}{8}: \\
\left(x^2+9 x+\frac{81}{4}\right)-10 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\fbox{$\frac{181}{8}$} \\
\end{array}
Step 9:
\begin{array}{l}
x^2+9 x+\frac{81}{4}=\left(x+\frac{9}{2}\right)^2: \\
\fbox{$\left(x+\frac{9}{2}\right)^2$}-10 \left(y^2-\frac{y}{2}+\frac{1}{16}\right)=\frac{181}{8} \\
\end{array}
Step 10:
\begin{array}{l}
y^2-\frac{y}{2}+\frac{1}{16}=\left(y-\frac{1}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & \left(x+\frac{9}{2}\right)^2-\text{10 }\fbox{$\left(y-\frac{1}{4}\right)^2$}=\frac{181}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $5 x^2+5 x-6 y^2+9 y-3=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-6 y^2+9 y+5 x^2+5 x-3=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Add }3 \text{to }\text{both }\text{sides}: \\
-6 y^2+9 y+5 x^2+5 x=3 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(5 x^2+5 x+\underline{\text{ }}\right)+\left(-6 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 4:
\begin{array}{l}
\left(5 x^2+5 x+\underline{\text{ }}\right)=5 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$5 \left(x^2+x+\underline{\text{ }}\right)$}+\left(-6 y^2+9 y+\underline{\text{ }}\right)=\underline{\text{ }}+3 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-6 y^2+9 y+\underline{\text{ }}\right)=-6 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right): \\
5 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$-6 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)$}=\underline{\text{ }}+3 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{5}{4}=\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
3+\frac{5}{4}=\frac{17}{4}: \\
5 \left(x^2+x+\frac{1}{4}\right)-6 \left(y^2-\frac{3 y}{2}+\underline{\text{ }}\right)=\fbox{$\frac{17}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-3}{2}}{2}\right)^2=\frac{9}{16} \text{on }\text{the }\text{left }\text{and }-6\times \frac{9}{16}=-\frac{27}{8} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{17}{4}-\frac{27}{8}=\frac{7}{8}: \\
5 \left(x^2+x+\frac{1}{4}\right)-6 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\fbox{$\frac{7}{8}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
5 \fbox{$\left(x+\frac{1}{2}\right)^2$}-6 \left(y^2-\frac{3 y}{2}+\frac{9}{16}\right)=\frac{7}{8} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{3 y}{2}+\frac{9}{16}=\left(y-\frac{3}{4}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 5 \left(x+\frac{1}{2}\right)^2-6 \fbox{$\left(y-\frac{3}{4}\right)^2$}=\frac{7}{8} \\
\end{array}
| khanacademy | amps |
Given the equation $6 x^2-7 x-5 y^2+y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-5 y^2+y+6 x^2-7 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-5 y^2+y+6 x^2-7 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(6 x^2-7 x+\underline{\text{ }}\right)+\left(-5 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(6 x^2-7 x+\underline{\text{ }}\right)=6 \left(x^2-\frac{7 x}{6}+\underline{\text{ }}\right): \\
\fbox{$6 \left(x^2-\frac{7 x}{6}+\underline{\text{ }}\right)$}+\left(-5 y^2+y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-5 y^2+y+\underline{\text{ }}\right)=-5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right): \\
6 \left(x^2-\frac{7 x}{6}+\underline{\text{ }}\right)+\fbox{$-5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }6 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-7}{6}}{2}\right)^2=\frac{49}{144} \text{on }\text{the }\text{left }\text{and }6\times \frac{49}{144}=\frac{49}{24} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
\frac{49}{24}-9=-\frac{167}{24}: \\
6 \left(x^2-\frac{7 x}{6}+\frac{49}{144}\right)-5 \left(y^2-\frac{y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{167}{24}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{-1}{5}}{2}\right)^2=\frac{1}{100} \text{on }\text{the }\text{left }\text{and }\frac{-5}{100}=-\frac{1}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{167}{24}-\frac{1}{20}=-\frac{841}{120}: \\
6 \left(x^2-\frac{7 x}{6}+\frac{49}{144}\right)-5 \left(y^2-\frac{y}{5}+\frac{1}{100}\right)=\fbox{$-\frac{841}{120}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2-\frac{7 x}{6}+\frac{49}{144}=\left(x-\frac{7}{12}\right)^2: \\
6 \fbox{$\left(x-\frac{7}{12}\right)^2$}-5 \left(y^2-\frac{y}{5}+\frac{1}{100}\right)=-\frac{841}{120} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-\frac{y}{5}+\frac{1}{100}=\left(y-\frac{1}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 6 \left(x-\frac{7}{12}\right)^2-5 \fbox{$\left(y-\frac{1}{10}\right)^2$}=-\frac{841}{120} \\
\end{array}
| khanacademy | amps |
Given the equation $-3 x^2-9 x-9 y^2+1=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-3 x^2-9 x+\left(1-9 y^2\right)=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }-9 y^2-3 x^2-9 x+1 \text{from }\text{both }\text{sides}: \\
3 x^2+9 x+\left(9 y^2-1\right)=0 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Add }1 \text{to }\text{both }\text{sides}: \\
9 y^2+3 x^2+9 x=1 \\
\end{array}
Step 4:
\begin{array}{l}
\text{Group }\text{terms }\text{together }\text{on }\text{the }\text{left }\text{hand }\text{side, }\text{with }\text{a }\text{placeholder }\text{constant}: \\
\left(3 x^2+9 x+\underline{\text{ }}\right)+9 y^2=\underline{\text{ }}+1 \\
\end{array}
Step 5:
\begin{array}{l}
\left(3 x^2+9 x+\underline{\text{ }}\right)=3 \left(x^2+3 x+\underline{\text{ }}\right): \\
\fbox{$3 \left(x^2+3 x+\underline{\text{ }}\right)$}+9 y^2=\underline{\text{ }}+1 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }3 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{3}{2}\right)^2=\frac{9}{4} \text{on }\text{the }\text{left }\text{and }3\times \frac{9}{4}=\frac{27}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
1+\frac{27}{4}=\frac{31}{4}: \\
3 \left(x^2+3 x+\frac{9}{4}\right)+9 y^2=\fbox{$\frac{31}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
x^2+3 x+\frac{9}{4}=\left(x+\frac{3}{2}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & 3 \fbox{$\left(x+\frac{3}{2}\right)^2$}+9 y^2=\frac{31}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-4 x^2-10 x-y^2+6 y+9=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
-y^2+6 y-4 x^2-10 x+9=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }9 \text{from }\text{both }\text{sides}: \\
-y^2+6 y-4 x^2-10 x=-9 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-4 x^2-10 x+\underline{\text{ }}\right)+\left(-y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-4 x^2-10 x+\underline{\text{ }}\right)=-4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right): \\
\fbox{$-4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)$}+\left(-y^2+6 y+\underline{\text{ }}\right)=\underline{\text{ }}-9 \\
\end{array}
Step 5:
\begin{array}{l}
\left(-y^2+6 y+\underline{\text{ }}\right)=-\left(y^2-6 y+\underline{\text{ }}\right): \\
-4 \left(x^2+\frac{5 x}{2}+\underline{\text{ }}\right)+\fbox{$-\left(y^2-6 y+\underline{\text{ }}\right)$}=\underline{\text{ }}-9 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-4 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{5}{2}}{2}\right)^2=\frac{25}{16} \text{on }\text{the }\text{left }\text{and }-4\times \frac{25}{16}=-\frac{25}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-9-\frac{25}{4}=-\frac{61}{4}: \\
-4 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)-\left(y^2-6 y+\underline{\text{ }}\right)=\fbox{$-\frac{61}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-1 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{-6}{2}\right)^2=9 \text{on }\text{the }\text{left }\text{and }-9=-9 \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
-\frac{61}{4}-9=-\frac{97}{4}: \\
-4 \left(x^2+\frac{5 x}{2}+\frac{25}{16}\right)-\left(y^2-6 y+9\right)=\fbox{$-\frac{97}{4}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+\frac{5 x}{2}+\frac{25}{16}=\left(x+\frac{5}{4}\right)^2: \\
-4 \fbox{$\left(x+\frac{5}{4}\right)^2$}-\left(y^2-6 y+9\right)=-\frac{97}{4} \\
\end{array}
Step 11:
\begin{array}{l}
y^2-6 y+9=(y-3)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -4 \left(x+\frac{5}{4}\right)^2-\fbox{$(y-3)^2$}=-\frac{97}{4} \\
\end{array}
| khanacademy | amps |
Given the equation $-5 x^2-5 x+5 y^2+3 y+10=0$, complete the square.
Step 1:
\begin{array}{l}
\begin{array}{l}
\text{Complete the square}: \\
5 y^2+3 y-5 x^2-5 x+10=0 \\
\end{array}
Step 2:
\begin{array}{l}
\text{Subtract }\text{10 }\text{from }\text{both }\text{sides}: \\
5 y^2+3 y-5 x^2-5 x=-10 \\
\end{array}
Step 3:
\begin{array}{l}
\text{Group }\text{terms }\text{with }x \text{and }y \text{separately, }\text{leaving }\text{placeholder }\text{constants}: \\
\left(-5 x^2-5 x+\underline{\text{ }}\right)+\left(5 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 4:
\begin{array}{l}
\left(-5 x^2-5 x+\underline{\text{ }}\right)=-5 \left(x^2+x+\underline{\text{ }}\right): \\
\fbox{$-5 \left(x^2+x+\underline{\text{ }}\right)$}+\left(5 y^2+3 y+\underline{\text{ }}\right)=\underline{\text{ }}-10 \\
\end{array}
Step 5:
\begin{array}{l}
\left(5 y^2+3 y+\underline{\text{ }}\right)=5 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right): \\
-5 \left(x^2+x+\underline{\text{ }}\right)+\fbox{$5 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right)$}=\underline{\text{ }}-10 \\
\end{array}
Step 6:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }x \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }-5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{1}{2}\right)^2=\frac{1}{4} \text{on }\text{the }\text{left }\text{and }\frac{-5}{4}=-\frac{5}{4} \text{on }\text{the }\text{right}: \\
\end{array}
Step 7:
\begin{array}{l}
-10-\frac{5}{4}=-\frac{45}{4}: \\
-5 \left(x^2+x+\frac{1}{4}\right)+5 \left(y^2+\frac{3 y}{5}+\underline{\text{ }}\right)=\fbox{$-\frac{45}{4}$} \\
\end{array}
Step 8:
\begin{array}{l}
\begin{array}{l}
\text{Take }\text{one }\text{half }\text{of }\text{the }\text{coefficient }\text{of }y \text{and }\text{square }\text{it. }\text{Then }\text{add }\text{it }\text{to }\text{both }\text{sides }\text{of }\text{the }\text{equation, }\text{multiplying }\text{by }\text{the }\text{factored }\text{constant }5 \text{on }\text{the }\text{right.} \\
\text{Insert }\left(\frac{\frac{3}{5}}{2}\right)^2=\frac{9}{100} \text{on }\text{the }\text{left }\text{and }5\times \frac{9}{100}=\frac{9}{20} \text{on }\text{the }\text{right}: \\
\end{array}
Step 9:
\begin{array}{l}
\frac{9}{20}-\frac{45}{4}=-\frac{54}{5}: \\
-5 \left(x^2+x+\frac{1}{4}\right)+5 \left(y^2+\frac{3 y}{5}+\frac{9}{100}\right)=\fbox{$-\frac{54}{5}$} \\
\end{array}
Step 10:
\begin{array}{l}
x^2+x+\frac{1}{4}=\left(x+\frac{1}{2}\right)^2: \\
-5 \fbox{$\left(x+\frac{1}{2}\right)^2$}+5 \left(y^2+\frac{3 y}{5}+\frac{9}{100}\right)=-\frac{54}{5} \\
\end{array}
Step 11:
\begin{array}{l}
y^2+\frac{3 y}{5}+\frac{9}{100}=\left(y+\frac{3}{10}\right)^2: \\
\fbox{$
\begin{array}{ll}
\text{Answer:} & \\
\text{} & -5 \left(x+\frac{1}{2}\right)^2+5 \fbox{$\left(y+\frac{3}{10}\right)^2$}=-\frac{54}{5} \\
\end{array}
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