Simplifying $\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}$

How do I simplify $\sqrt{(4+2\sqrt{3})}+\sqrt{(4-2\sqrt{3})}$?

I've tried to make it $x$ and square both sides but I got something extremely complicated and it didn't look right.

I got $2\sqrt{3}$ on wolframalpha, but I'm not sure how is it possible?

Help would be appreciated! Thanks!

\begin{align} &\ \ \ \sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}} \\ \\ &=\sqrt{(\sqrt{3}+1)^2}+\sqrt{(\sqrt{3}-1)^2}\ \\ \\ &=\sqrt{3}+1+\sqrt{3}-1 \\ \\ &=\boxed{2\sqrt{3}} \end{align}

You have the right idea in squaring and then taking the square root.

Note that $k = (\sqrt{4 + 2\sqrt{3}} + \sqrt{4 - 2\sqrt{3}})^{2} = 4 + 2\sqrt{3} + 2\sqrt{(4+2\sqrt{3})(4-2\sqrt{3})} + 4 - 2\sqrt{3}$

But this is just:

$k = 8 + 2\sqrt{16 - 12} = 12$

So

$\sqrt{k} = 2\sqrt{3}$