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.

You can rewrite this equation as :
$$\left(x+\frac{y}{2}\right)^2+\frac{3y^2}{4}+y\sqrt{3}+1=0$$
which can be transformed further into :
$$\left(x+\frac{y}{2}\right)^2+\frac{3}{4}\left(y^2+\frac{4y}{\sqrt3}+\frac{4}{3}\right)=0$$
Finally, you should notice that :
$$y^2+\frac{4y}{\sqrt3}+\frac{4}{3}=\left(y+\frac{2}{\sqrt3}\right)^2$$
When the sum of two positive numbers is zero, each of them must be $0$.
Hence :
$$x+\frac{y}{2}=0\qquad\mathrm{and}\qquad y+\frac{2}{\sqrt3}=0$$
which leads to :
$$x=\frac{1}{\sqrt3}\qquad\mathrm\qquad y=-\frac{2}{\sqrt3}$$
Finally :
$$\boxed{xy=-\frac{2}{3}}$$
