<s>[INST] 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!
 [/INST] Since $(4+2\sqrt3)(4-2\sqrt3)=16-12=4$, try squaring:
$$
\begin{align}
\left(\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}\right)^2
&=(4+2\sqrt3)+(4-2\sqrt3)+2\sqrt{(4+2\sqrt3)(4-2\sqrt3)}\\
&=8+2\sqrt{16-12}\\[6pt]
&=12
\end{align}
$$
Therefore, $\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}=2\sqrt3$
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