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

\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

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

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, $\sqrt{4+2\sqrt3}+\sqrt{4-2\sqrt3}=2\sqrt3$