Which of the following is equal to $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$?
$\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6$
We find the answer by squaring, then square rooting the expression.
\begin{align*} &\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\\\\ = \ &\sqrt{\left(\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\right)^2}\\\\ = \ &\sqrt{9-6\sqrt{2}+2\sqrt{(9-6\sqrt{2})(9+6\sqrt{2})}+9+6\sqrt{2}}\\\\ = \ &\sqrt{18+2\sqrt{(9-6\sqrt{2})(9+6\sqrt{2})}}\\\\ = \ &\sqrt{18+2\sqrt{9^2-(6\sqrt{2})^2}}\\\\ = \ &\sqrt{18+2\sqrt{81-72}}\\\\ = \ &\sqrt{18+2\sqrt{9}}\\\\ = \ &\sqrt{18+6}\\\\= \ &\sqrt{24}\\\\ = \ &\boxed{2\sqrt{6} \ \mathbf{(B)}}. \end{align*}