1. ## Completing the square

How do I complete the square of x^2 + xy + y^2 = 3?

2. ## Re: Completing the square

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

That's easy:

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

$\left(y+\dfrac{x}{2}\right)^2 = 3\left(1-\dfrac{x^2}{4}\right)$

3. ## 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.

4. ## Re: Completing the square

I think this version is clearly superior!

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

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

$\left(x + \dfrac y 2\right)^2 = 3\left(1 - \dfrac {y^2} 4\right)$

5. ## 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 $(x,y,u) = (-\sqrt{3},\sqrt{3},2)$

6. ## Re: Completing the square

Thank you so much!!