<s>[INST] Simplifying nested square roots ($\sqrt{6-4\sqrt{2}} + \sqrt{2}$) I guess I learned it many years ago at school, but I must have forgotten it. From a geometry puzzle I got to the solution
$\sqrt{6-4\sqrt{2}} + \sqrt{2}$
My calculator tells me that (within its precision) the result equals exactly 2, but I have no idea how to transform the calculation to symbolically get to that result.
(I can factor out one $\sqrt{2}$ from both terms, but that does not lead me anywhere, either)
 [/INST] Since 
\begin{align}
6-4\sqrt{2}&=4-2\cdot 2\cdot\sqrt{2}+2\\
&=(2-\sqrt{2})^2
\end{align}
Where $2-\sqrt{2}>0$, it follows 
$\sqrt{6-4\sqrt{2}}=2-\sqrt{2}$.
Then $$\sqrt{6-4\sqrt{2}}+\sqrt{2}=\color{blue}{2}$$
</s>