$\displaystyle x^2-8x+y^2+2y=-14$

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- Oct 21st 2012, 07:59 PMEraser147Can anyone show me how to complete the square on this
$\displaystyle x^2-8x+y^2+2y=-14$

- Oct 21st 2012, 08:06 PMProve ItRe: Can anyone show me how to complete the square on this
- Oct 21st 2012, 08:17 PMEraser147Re: Can anyone show me how to complete the square on this
It would be $\displaystyle (x-4)^2-2+(y+1)^2+13$. Which is wrong. Solutions manual says it's supposed to be $\displaystyle (x-4)^2-16+(y+1)^2-1$

- Oct 21st 2012, 08:25 PMProve ItRe: Can anyone show me how to complete the square on this
Remember that this is an EQUATION.

- Oct 21st 2012, 08:27 PMEraser147Re: Can anyone show me how to complete the square on this
so would it be what i had = 0 or = -14. I am so confused.

- Oct 21st 2012, 08:32 PMProve ItRe: Can anyone show me how to complete the square on this
Start with your initial equation, then whatever you add to complete the square on the x terms and the y terms, add to the other side as well, to keep the equation balanced.

- Oct 21st 2012, 08:34 PMEraser147Re: Can anyone show me how to complete the square on this
Can you show me? I still don't quite follow.

- Oct 21st 2012, 08:49 PMProve ItRe: Can anyone show me how to complete the square on this
You can show me every step you've taken, then I can guide you where you have gone wrong...

- Oct 21st 2012, 08:57 PMEraser147Re: Can anyone show me how to complete the square on this
okay this is what I did, I separated the x and y and completed the square for both x and y terms so I resulted in this $\displaystyle (x-4)^2-2+(y+1)^2+13$. I took $\displaystyle x^2-8x+14$ and compeleted the square first and I did the same by adding 14 on the y terms as well.

- Oct 22nd 2012, 10:08 AMSorobanRe: Can anyone show me how to complete the square on this
Hello, Eraser147!

Quote:

$\displaystyle \text{Complete the square: }\:x^2-8x+y^2+2y\:=\:\text{-}14$

We have: .$\displaystyle x^2 - 8x \qquad + y^2 + 2y \qquad \;=\;\text{-}14$

Then:. . . . $\displaystyle x^2 - 8x {\color{red}\:+\:16} + y^2 + 2y {\color{green}\:+\:1} \;=\;\text{-}14 {\color{red}\:+\:16} {\color{green}\:+\:1} $

Therefore: . . . . . .$\displaystyle (x-4)^2 + (y+1)^2 \;=\;3$

- Oct 22nd 2012, 05:21 PMEraser147Re: Can anyone show me how to complete the square on this
Thanks Soroban, you are the best.