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Math Help - Completing the square

  1. #1
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    Completing the square

    How do I complete the square of x^2 + xy + y^2 = 3?
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  2. #2
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    Re: Completing the square

    What are you trying to do? Solve for $y$ in terms of $x$?

    That's easy:

    $y^2+xy+ \dfrac{x^2}{4} = \dfrac{x^2}{4} + 3-x^2 = 3 - \dfrac{3x^2}{4}$

    $\left(y+\dfrac{x}{2}\right)^2 = 3\left(1-\dfrac{x^2}{4}\right)$
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  3. #3
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    Re: Completing the square

    Ive differentiated to find three equations

    2x - 2ux - uy = 0
    2y - ux - 2uy = 0
    -(x^2 + xy + y^2 - 3) = 0

    and need to find all solutions of x,y, and u. Ive solved for u and found x, y, u, are (1,1, 2/3) (-1,-1,2/3) but cannot find the other two solutions. Any help with that would be much appreciated, thought it had to do with completing the square but I guess not.
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  4. #4
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    Re: Completing the square

    I think this version is clearly superior!

    $x^2 + x y + y^2 = 3$

    $\left(x + \dfrac y 2\right)^2 + \dfrac {3 y^2} 4 = 3$

    $\left(x + \dfrac y 2\right)^2 = 3\left(1 - \dfrac {y^2} 4\right)$
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  5. #5
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    Re: Completing the square

    Solving for $x$ in the first equation gives $x = \dfrac{uy}{2(1-u)}$

    Solving for $x$ in the second equation gives: $x = \dfrac{2y(1-u)}{u}$

    Setting them equal gives $u^2y = 4y(1-u)^2$. So, if $y=0$, then $x=0$, which is not a solution to the third equation. Hence $y \neq 0$, allowing you to divide both sides by $y$. Taking the square root gives:

    $u = \pm 2(1-u)$

    $u = 2-2u$ or $u = 2u-2$ gives $u = 2$ or $u = \dfrac{2}{3}$.

    If $u = \dfrac{2}{3}$, you found the two solutions. So, try $u = 2$:

    If $u=2$, then $x=-y$ in both the first and second equations. Plugging this into the third equation:

    $(-y)^2 +(-y)y + y^2-3 = y^2-y^2+y^2-3 = y^2-3 = 0$, so $y = \pm \sqrt{3}$. Hence, the other two answers are:

    $(x,y,u) = (\sqrt{3},-\sqrt{3},2)$ or $(x,y,u) = (-\sqrt{3},\sqrt{3},2)$
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  6. #6
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    Re: Completing the square

    Thank you so much!!
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