<s>[INST] if $x \neq 0$ and $x = \sqrt{4xy - 4y^2}$, then how does expressing $x$ in terms of $y$ mean $x = 2y$? I have the equation $x=\sqrt{4xy - 4y^2}$, and I know that $x=2y$ when expressed in terms of $y$, but I'm not sure of the process to get there. 
I know that 
\begin{align}\sqrt{4xy - 4y^2} &= \sqrt{4y(x-y)}\\
&= 2 \sqrt{y(x-y)} \\
&= 2(xy-y^2)^{\frac12}
\end{align} 
but pretty stumped.
 [/INST] $x =\sqrt{4xy + 4y^2}$
$x^2 = 4xy - 4y^2$
$x^2 - 4xy + 4y^2 = 0$
$x = \frac{4y \pm \sqrt{16y^2 - 16y^2}}{2}$
$x = \frac{4y \pm \sqrt{0}}{2}$
$x = 2y$
</s>