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)
 A: Start by trying to simplify $\sqrt{6-4\sqrt{2}}$. Let's assume there is some number $p+q\sqrt{2}$ for which
$$\sqrt{6-4\sqrt{2}} = p + q\sqrt{2}$$
Squaring both sides gives 
$$6-4\sqrt{2} = (p + q\sqrt{2})^2 = p^2+2q^2 + 2pq\sqrt{2}$$
Comparing coefficients gives $6=p^2+2q^2$ and $-4=2pq$, i.e. $-2=pq$.
We need to solve $p^2+2q^2=6$ and $pq = -2$ simultaneously. 
If $pq=-2$ then $q=-\frac{2}{p}$ and we can substitute this into $p^2+2q^2=6$. We get
\begin{eqnarray*}
p^2+2q^2 &=& 6 \\ \\
p^2 + 2\left(-\frac{2}{p}\right)^2 &=& 6 \\ \\
p^2 + \frac{8}{p^2} &=& 6 \\ \\
p^4+8 &=& 6p^2 \\ \\
p^4-6p^2+8 &=& 0 \\ \\
(p^2-2)(p^2-4) &=& 0
\end{eqnarray*}
Either $p^2-2=0$ or $p^2-4=0$, i.e. $p=\pm\sqrt{2}$ or $p=\pm 2$. The only valid solutions are $p = \pm2$ because we usually assume that $p$ and $q$ are rational numbers, i.e. fractions. 
If $p=\pm 2$ then $pq=2$ gives $\pm 2q=-2$, and so $q=\mp 1$. Hence
$$p+q\sqrt{2} = \pm(2-\sqrt{2})$$
Recall that $\sqrt{6-4\sqrt{2}} = p + q\sqrt{2}$ and since, by definition, $\sqrt{6-4\sqrt{2}} \ge 0$ we conclude
$$\sqrt{6-4\sqrt{2}} = 2-\sqrt{2}$$
Finally:
$$\sqrt{6-4\sqrt{2}} \ \ {\color{red}{+\sqrt{2}}}= 2-\sqrt{2} \ \ {\color{red}{+\sqrt{2}}} = 2$$
A: 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 $$\sqrt{6-4\sqrt{2}}+\sqrt{2}=\color{blue}{2}$$
A: 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 $6-4\sqrt2=2^2-2\cdot2\cdot\sqrt2+(\sqrt2)^2$.
A: $$\sqrt{6-4\sqrt{2}}+\sqrt{2}=\sqrt{6-\sqrt{4^2}\sqrt{2}}+\sqrt{2}=$$
$$\sqrt{6-\sqrt{16}\sqrt{2}}+\sqrt{2}=\sqrt{6-\sqrt{16\cdot2}}+\sqrt{2}=$$
$$\sqrt{6-\sqrt{32}}+\sqrt{2}=\sqrt{\sqrt{6^2}-\sqrt{32}}+\sqrt{2}=$$
$$\sqrt{\sqrt{36}-\sqrt{32}}+\sqrt{2}=\sqrt{\sqrt{4\cdot9}-\sqrt{4\cdot8}}+\sqrt{2}=$$
$$\sqrt{2\sqrt{9}-2\sqrt{8}}+\sqrt{2}=\sqrt{2\left(\sqrt{9}-\sqrt{8}\right)}+\sqrt{2}=$$
$$\sqrt{2}\sqrt{\sqrt{9}-\sqrt{8}}+\sqrt{2}=\sqrt{2}\left(\sqrt{\sqrt{9}-\sqrt{8}}+1\right)=$$
$$\sqrt{2}\left(\sqrt{3-2\sqrt{2}}+1\right)=\sqrt{2}\left(\sqrt{2\left(\frac{3}{2}-\sqrt{2}\right)}+1\right)=$$
$$\sqrt{2}\left(\sqrt{2}\sqrt{\frac{3}{2}-\sqrt{2}}+1\right)=
\sqrt{2}\left(\sqrt{2}\left(1-\frac{1}{\sqrt{2}}\right)+1\right)=$$
$$\sqrt{2}\left(\sqrt{2}-\frac{\sqrt{2}}{\sqrt{2}}+1\right)=\sqrt{2}\left(\sqrt{2}-1+1\right)=$$
$$\sqrt{2}\left(\sqrt{2}\right)=\left(\sqrt{2}\right)^2=2$$

EDIT:
$$\sqrt{\frac{3}{2}-\sqrt{2}}=\sqrt{\frac{3}{2}-\frac{2\sqrt{2}}{2}}=$$
$$\sqrt{\frac{3-2\sqrt{2}}{2}}=\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{4}}=$$
$$\frac{\sqrt{\left(2-\sqrt{2}\right)^2}}{\sqrt{4}}=\frac{2-\sqrt{2}}{2}=1-\frac{\sqrt{2}}{2}=1-\frac{1}{\sqrt{2}}$$
