Evaluate this square root $\sqrt{6 + 2\sqrt{5}} + \sqrt{6 - 2\sqrt{5}}$ 
I have no clue where to begin. I would appreciate a hint, the answer should be
$2\sqrt{5}$
In general, how do you evaluate 
$\sqrt{a + b} + \sqrt{a - b}$?
Thanks!

Set $r=\sqrt{6 + 2\sqrt{5}} + \sqrt{6 - 2\sqrt{5}}$ and observe that $r>0$; then
\begin{align}
r^2
&=6+2\sqrt{5}+2\sqrt{(6 + 2\sqrt{5})(6 - 2\sqrt{5})} + 6 - 2\sqrt{5}\\
&=12+2\sqrt{36-20}\\
&=12+2\sqrt{16}\\
&=12+8=20
\end{align}
Thus $r=\sqrt{r^2}=\sqrt{20}=2\sqrt{5}$.
More generally, if $a>b$ and $r=\sqrt{a+b}+\sqrt{a-b}$, then
$$
r^2=2a+2\sqrt{a^2-b^2}
$$
which can be simplified if $a^2-b^2$ is a perfect square.

There is the general formula
$$
\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}
$$
with its companion
$$
\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}}
$$
(supposing $a>0$ and $a^2-b\ge0$), but it's in general better to look for a perfect square; in $6+2\sqrt{5}$ you have a double product $2\sqrt{5}$, so writing $6=1+5$ seems the best:
$$
6+2\sqrt{5}=1+2\sqrt{5}+(\sqrt{5})^2=(1+\sqrt{5})^2
$$
