2011 AMC 10A Problem 16

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

A sum of two conjugate-looking radicals is tamed by squaring: the awkward parts cancel in the plain sum, and the cross term involves the product of conjugates, which is rational. The only care needed is the sign when taking the final square root, which is fixed by noting $S$ is positive.

Denest each radical. Seek $\sqrt{9 - 6\sqrt2} = \sqrt a - \sqrt b$ with $a + b = 9$ and $2\sqrt{ab} = 6\sqrt2$, i.e. $ab = 18$: $a = 6$, $b = 3$. Then $\sqrt{9 - 6\sqrt2} = \sqrt6 - \sqrt3$ and $\sqrt{9 + 6\sqrt2} = \sqrt6 + \sqrt3$, whose sum is $2\sqrt6$.

• Estimating numerically: the sum is about 4.90, but choice (C) is about 4.95 and choice (E) is 6, so a rough decimal check cannot separate (B) from (C).
• Writing $\sqrt{9 - 6\sqrt2} = 3 - \sqrt{6\sqrt2}$, splitting a root over a difference, which gives the false answer $6$.