How to equalize correctly? If i have this number:
$2 \sqrt{2-\sqrt{3}}$ and i want to find some $x,y$ nonzero real numbers such that $2\sqrt{2-\sqrt{3}} = \sqrt{x} + \sqrt{y}$ 
And for that, i do this:
$(2 \sqrt{2-\sqrt{3}})^2 = x + 2\sqrt{xy} + y$
$4(2-\sqrt{3})=(x+y)+2\sqrt{xy}$
$(8)+(-4\sqrt{3})=(x+y)+(2\sqrt{xy})$
Then:
$i) 8 = x+y$,
$ii)-4\sqrt{3} = 2\sqrt{xy} => -2\sqrt{3}=\sqrt{xy}$
$ii) = (-2\sqrt{3})^2= (\sqrt{xy})^2 => 4\cdot 3=xy , x = 12/y$
And solving the equation $y^2-8y+12=0$ gives $y_{1,2} = \{6,2\}$
But $2\sqrt{2-\sqrt{3}} \neq \sqrt{6} + \sqrt{2}$
I know  that the correct value must be $\sqrt{6} - \sqrt{2}$ but my result is different. What is wrong with my development? 
 A: You should assume a binomial of the form whose root you desire. In this case, you should have supposed the root is $$\sqrt x -\sqrt y$$ instead.
To be specific, the problem in the above calculation is in your step ii,  where you set $$-\sqrt{12}=\sqrt{xy}.$$ But this is impossible if you're dealing only with real numbers. It seems you need to note that the symbol $\sqrt{}$ denotes a function which, by definition, assumes nonnegative values. Thus, you can see that your equation is false, for it says a negative number is equal to a nonnegative one. That's a contradiction.
A: I can write $2\sqrt{2-\sqrt{3}} = \sqrt{y}+\sqrt{x}$ as: $2\sqrt{(\frac{\sqrt3}{\sqrt2}-\frac{\sqrt2}{2})^2} = \sqrt{y}+\sqrt{x}=2(\frac{\sqrt3}{\sqrt2}-\frac{\sqrt2}{2})$ or $2\sqrt{(\frac{1}{\sqrt2}-\frac{\sqrt6}{2})^2} = \sqrt{y}+\sqrt{x}=2(\frac{1}{\sqrt2}-\frac{\sqrt6}{2})$. This can be easily obtained by the system:
$$\left\{\begin{matrix}
a^2+b^2=2
\\2ab=-\sqrt{3}
\end{matrix}\right.$$
So: $\sqrt{x}=-\frac{\sqrt2}{2}$, but this is impossibile because the square root can't be negative: same arguments for $\sqrt{y}$.
