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!

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