2011 AMC 10A Problem 16

Call the sum $S$. Each square root on its own is messy, but squaring a sum of two square roots makes most of the mess cancel, so square $S$.

1. Check the signs. $6\sqrt2$ is about $8.49$, which is less than $9$, so both numbers under the roots are positive. Both roots are positive, so $S$ is positive.
2. Square using $(p + q)^2 = p^2 + q^2 + 2pq$. Here $p = \sqrt{9 - 6\sqrt2}$ and $q = \sqrt{9 + 6\sqrt2}$.
- $p^2 + q^2 = (9 - 6\sqrt2) + (9 + 6\sqrt2) = 18$. The $6\sqrt2$ parts cancel.
- $pq = \sqrt{(9 - 6\sqrt2)(9 + 6\sqrt2)}$. This is a "difference of squares": $(9 - 6\sqrt2)(9 + 6\sqrt2) = 9^2 - (6\sqrt2)^2 = 81 - 72 = 9$. So $pq = \sqrt9 = 3$.
3. So $S^2 = 18 + 2 \times 3 = 24$.
4. $S$ is positive, so $S = \sqrt{24} = \sqrt{4 \times 6} = 2\sqrt6$.

The answer is $\boxed{\textbf{(B)}\ 2\sqrt6}$.

When you add two square roots whose insides differ only by a plus or minus sign, square the sum. The plus and minus parts cancel in $p^2 + q^2$, and the product $pq$ becomes a difference of squares, which is a whole number. At the end, take the positive square root, because the original sum was positive.

Rewrite each root as a simpler expression. Look for numbers with $(\sqrt6 - \sqrt3)^2 = 6 + 3 - 2\sqrt{18} = 9 - 2 \cdot 3\sqrt2 = 9 - 6\sqrt2$. So $\sqrt{9 - 6\sqrt2} = \sqrt6 - \sqrt3$ (this is positive, since $\sqrt6 > \sqrt3$). In the same way $\sqrt{9 + 6\sqrt2} = \sqrt6 + \sqrt3$. Adding, the $\sqrt3$ parts cancel and the sum is $2\sqrt6$.

Let $S = \sqrt{9 - 6\sqrt2} + \sqrt{9 + 6\sqrt2}$. Both radicands are positive ($6\sqrt2 \approx 8.49 < 9$), and $S > 0$.

Square it:
$$S^2 = (9 - 6\sqrt2) + (9 + 6\sqrt2) + 2\sqrt{(9 - 6\sqrt2)(9 + 6\sqrt2)}.$$
The product under the last root is a difference of squares:
$$(9 - 6\sqrt2)(9 + 6\sqrt2) = 81 - 72 = 9,$$
so the cross term is $2\sqrt9 = 6$. Therefore $S^2 = 18 + 6 = 24$, and since $S > 0$,
$$S = \sqrt{24} = 2\sqrt6.$$

The answer is $\boxed{\textbf{(B)}\ 2\sqrt6}$.