How to factorize this $\sqrt{8 - 2\sqrt{7}}$? When I was at high school, our teacher showed us a technique to simplify square roots like this  $\sqrt{8 - 2\sqrt{7}}$ that I forgot.
It was something like 8 = 7+1; 7 = 7*1; and using them we could represent $\sqrt{8 - 2\sqrt{7}}$ in simpler form. I would be happy if you can show how it works, and how this technique is called.

You could approach this by setting $\sqrt{8 - 2\sqrt{7}} = \sqrt a \pm \sqrt b$ for some $a,b$. Then, squaring both sides, we have:
$$8 - 2\sqrt{7} = a \pm 2\sqrt{ab} + b,$$
so that $8=a+b$ and $-2\sqrt7 = \pm 2\sqrt{ab}$.
In other words, the $\pm$ sign must be $-$, and we now have a system of two equations $\{8=a+b, 7=ab\}$ which is readily solved. Either $a=7$ and $b=1$ (which works), or $a=1$ and $b=7$ (which we reject because $1-\sqrt 7 < 0$).
