How does $\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$ become $\sqrt{2(2+\sqrt{2})}$? I'd like to know how can one simplify the following expression
$$\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}}$$
into
$$\sqrt{2(2+\sqrt{2})}.$$
Wolfram alpha suggests it as an alternative form, and numerically it's easy to verify, but I can't find the right algebra to show they are indeed equivalent.
Note I ran into this problem, trying to do: $2\cos(\pi/8)+2\sin(\pi/8)$, where
$$2\cos(\pi/8)=\sqrt{2+\sqrt{2}},$$
$$2\sin(\pi/8)=\sqrt{2-\sqrt{2}}.$$

Following @Yousef's answer:

 \begin{align} \sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}} &= x \\ \left(\sqrt{2+\sqrt{2}}+\sqrt{2-\sqrt{2}} \right)^2 &= x^2 \\ 2+\sqrt{2}+2-\sqrt{2}+2\sqrt{4-2} &= x^2 \\ 4+2\sqrt{2} &= x^2 \implies x=\sqrt{4+2\sqrt{2}} \end{align} as $x>0$.

