Technique for simplifying, e.g. $\sqrt{ 8 - 4\sqrt{3}}$ to $\sqrt{6} - \sqrt{2}$ How to find the square root of an irrational expression, to simplify that root. e.g.:
$$
\sqrt{ 8 - 4\sqrt{3} } = \sqrt{6} - \sqrt{2}
$$
Easy to verify:
\begin{align}
(\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} +2 
 = 8 - 4 \sqrt{3}
\end{align}
But how to work it out in the first place? I feel there's a standard technique (Completing-the-square? Quadratic formula?), but don't recall it or what it's called...
BTW: this came up in verifying equivalence of different calculations of $\cos{75°}$ (the above divided by $4$), as $\cos{\frac{90°+60°}{2}}$ vs $\cos{(45°+30°)}$, from 3Blue1Brown's lockdown video on complex numbers and trigonometry.

I don't think there's a name for this procedure but let's apply it to $\sqrt{8-4\sqrt3}$.
If you suspect this equals $\sqrt a\pm\sqrt b$ with rationals $a$ and $b$, then
$$8-4\sqrt3=(\sqrt a\pm\sqrt b)^2=(a+b)\pm2\sqrt{ab}$$
so you want to solve simultaneously $a+b=8$ and $-4\sqrt{3}=\pm2\sqrt{ab}$.
So you need the minus sign, and $ab=12$. Then $a$ and $b$ are roots of the
quadratic equation
$$(X-a)(X-b)=X^2-(a+b)X+ab=X^2-8X+12.$$
This does has rational roots: $2$ and $6$.
