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.
 A: Multiply both sides on 4
$$x^2+xy+y^2+ \sqrt{3}y + 1=0$$
$$4x^2+4xy+y^2+3y^2+ 4\sqrt{3}y + 4=0$$
$$(2x+y)^2+(\sqrt3y+2)^2=0$$
Then $y=-\frac2{\sqrt3}, x=\frac1{\sqrt3}$
$$xy=-\frac2{3}$$
A: 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 :
$$\boxed{xy=-\frac{2}{3}}$$
A: Setting $t=xy\in\mathbb R$, the equation becomes (knowing that $y\ne0$):
$$\frac{t^2}{y^2}+t+y^2+\sqrt3y+1=0.$$
The discriminant of this quadratic equation is $$\Delta=b^2-4ac=\frac{-3y^2-4\sqrt3y-4}{y^2}=-3\left(1-\dfrac2{\sqrt3y}\right)^2$$ and for a real solution to exist, it must be zero.
Then 
$$xy=t=-\frac b{2a}=-\frac{y^2}2=-\frac23.$$
A: Here we have two variables. Looks impossible. So our approach is pretty obvious. Its one of those kinds of problem where there equation holds only for a particular value of x and y. 
$ (x + (y/2))^2 + ((√3/2)y +1)^2 =0 $ .

Two non negative terms (squares) adding to zero implies both 0. 
Therefore $ x=-y/2$ and $ y= -(2/√3) $
A: 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 $\Delta=0$)
$$y_{1,2}=\frac{-(x+\sqrt{3})}{2}=-\frac{2}{\sqrt{3}}$$
