<s>[INST] 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.
 [/INST] If I recall correctly, you make the assumption that your expression takes the form $\sqrt{a}\pm\sqrt{b}$:
$\sqrt{8-4\sqrt{3}} = \sqrt{a}\pm\sqrt{b}$
$8-4\sqrt{3} = 8-\sqrt{48} = a\pm2\sqrt{ab}+b$
We can see that the irrational part must be assigned the negative sign.
Equating rational and irrational parts:
$a+b = 8$
$-2\sqrt{ab} = -\sqrt{48} \implies ab = 12$ 
Then $a = \frac{12}{b} \implies \frac{12}{b}+b = 8 \implies 12 + b^2 = 8b \implies b^2 - 8b + 12 = (b-6)(b-2) = 0$
Since we know the answer is positive, take b = 2 and a = 6.
$\sqrt{8-4\sqrt{3}} = \sqrt{6}-\sqrt{2}$
</s>