<s>[INST] Solve $x^2+xy+y^2+ \sqrt{3}y + 1=0$ for $xy$. I have a problem on my textbook:

$x,y \in\mathbb{R}$ and  $x^2+xy+y^2+ \sqrt{3}y + 1=0$ is given. Find the value of $xy$. 

I couldn't find.
 [/INST] Lets group the terms so we get a quadratic in $y$
$$y^2+(x+\sqrt{3})y+x^2+1=0$$
Now for a solution to exist we must have that the discriminant $\Delta\geq 0$
$$\Delta=(x+\sqrt{3})^2-4(x^2+1)=-3x^2+2\sqrt{3}-1=-(3x^2-2\sqrt{3}+1)=-(\sqrt{3}x-1)^2$$
Since a square is always $\geq 0$ we have that $\Delta\leq0$ (because of the minus) which implies $\sqrt{3}x-1=0$,$x=\frac{1}{\sqrt{3}}$
From the quadratic formula we have(since $\Delta=0$)
$$y_{1,2}=\frac{-(x+\sqrt{3})}{2}=-\frac{2}{\sqrt{3}}$$
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