# 2011 AMC 10A Problems/Problem 16

## Problem 16

Which of the following is equal to $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$? $\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6$

## Solution 1 (Bash)

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*}

## Solution 2 (FASTER!)

We can change the insides of the square root into a perfect square and then simplify. $$\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$$ $$= \sqrt{6-6\sqrt{2}+3}+\sqrt{6+6\sqrt{2}+3}$$ $$= \sqrt{\left(\sqrt{6}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}$$ $$= \sqrt{6}-\sqrt{3}+\sqrt{6}+\sqrt{3}$$ $$= \boxed{B) 2\sqrt{6}}$$

## Solution 3 (FASTEST)

Square roots remind us of squares. So lets try to make $9 - 6\sqrt{2} = (a-b)^2$. Doing a little experimentation we find that $$9 - 6\sqrt{2} = (\sqrt{6} - \sqrt{3})^2.$$ Similarly since $9 + 6\sqrt{2} = (a+b)^2$ we know that $$9 + 6\sqrt{2} = (\sqrt{6} + \sqrt{3})^2.$$

We want to find $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$. Using what we found above we know $$\sqrt{9 -6\sqrt{2}}+\sqrt{9+6\sqrt{2}} = (\sqrt{6} - \sqrt{3}) + (\sqrt{6} + \sqrt{3}) = 2\sqrt{6}.$$ This is nothing but $\boxed{B) 2\sqrt{6}}$.

~coolmath_2018

Note: This is basically just Solution 2 except you "do a little experimentation"

## Video Solution

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 