How do I simplify $\sqrt {4(2- \sqrt{3})}$ into $\sqrt{6} - \sqrt{2}$ This might be a stupid question, but how do I get from $$\sqrt {4(2- \sqrt{3})}$$ to $$\sqrt{6} - \sqrt{2}$$ It is obvious if you squared both, they both equal $8 - 4 \sqrt{3}$, but I'm wondering how you can find the answer from the original expression.
 A: Such roots can  be computed by a Simple Denesting Rule:. $ $  The radicand is $\,  8-\sqrt{48} = 4(2-\sqrt3)$
Here $\ 8-\sqrt{48}\ $ has norm $= 16.\:$ $\rm\ \color{blue}{Subtracting\ out}\,\ \sqrt{norm}\ = 4\,\ $ yields $\,\  4-\sqrt{48} $
which has $\, {\rm\ \sqrt{trace}}\, =\, \sqrt{8},\,\ \ \rm so\ \ \  \color{brown}{Dividing\ it\ out}\,\ $ of the above  yields $\ \dfrac{\sqrt{16}-\sqrt{48}}{\sqrt{8}} =\!\!\!\underbrace{\sqrt{2}-\sqrt 6}_{\rm negate\ for\ root > 0}$
A: Suppose $8-4\sqrt 3= (a-b\sqrt 3)^2$
This gives $a^2+3b^2=8$ and $2ab=4$ so that $ab=2$
This means that $a^2b^2+3b^4=8b^2$ or $4+3b^4=8b^2$ or $$3b^4-8b^2+4=0=(3b^2-2)(b^2-2)$$
So either $b^2=2$ or $b^2=\frac 23$.
We have also $a^2b^2=4$ so that $a^2=2$ or $a^2=6$
This should suffice to identify the solution which corresponds to the principal value (non-negative real for square root of non-negative real) of the square root.
Since the two square roots in the original are both positive, we have a unique solution. But the alternate possible signs indicate that a quartic is in the background. The other solutions of the quartic correspond to alternative choices of sign for the square root.
A: Note that 
\begin{align}
4(2-\sqrt{3}) = 8 - 4\sqrt{3} = (\sqrt{6})^2 - 2\sqrt{6}\sqrt{2} + (\sqrt{2})^2
=
(\sqrt{6}-\sqrt{2})^2.
\end{align}
Therefore,
$$
\sqrt{6} -\sqrt{2} = \sqrt{4(2-\sqrt{3})}.
$$
