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Thread: Can anyone show me how to complete the square on this

  1. #1
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    Can anyone show me how to complete the square on this

    $\displaystyle x^2-8x+y^2+2y=-14$
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    Re: Can anyone show me how to complete the square on this

    Quote Originally Posted by Eraser147 View Post
    $\displaystyle x^2-8x+y^2+2y=-14$
    Treat the x terms and the y terms separately. Complete the square on each.
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    Re: 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 $\displaystyle (x-4)^2-16+(y+1)^2-1$
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    Re: Can anyone show me how to complete the square on this

    Remember that this is an EQUATION.
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    Re: 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.
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    Re: 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.
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    Re: Can anyone show me how to complete the square on this

    Can you show me? I still don't quite follow.
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    Re: 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...
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    Re: 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.
    Last edited by Eraser147; Oct 21st 2012 at 09:01 PM.
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    Re: Can anyone show me how to complete the square on this

    Hello, Eraser147!

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

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

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    Re: Can anyone show me how to complete the square on this

    Thanks Soroban, you are the best.
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