<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] You want to calculate $a+b$ where $a^2=4+2\sqrt3$ and $b^2=4-2\sqrt3$.
You could notice that $a^2$ and $b^2$ are roots of the quadratic equation
$$x^2-8x+4=0.$$
(To find this equation you just need to find appropriate coefficients in quadratic formula.)
Now from Viete's formula you get
\begin{align*}
a^2+b^2&=8\\
a^2b^2&=4
\end{align*}
which (together with the fact that $a,b>0$) leaves you with $ab=2$ and $$(a+b)^2=a^2+b^2+2ab=8+2\cdot3=12.$$
You get that $$a+b=\sqrt{12}=2\sqrt3.$$
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