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)

Hint:
If the expression under the first radical is a perfect square, the double product $4\sqrt2$ factors as $2\cdot2\cdot\sqrt2$. Then you indeed have $6-4\sqrt2=2^2-2\cdot2\cdot\sqrt2+(\sqrt2)^2$.
