Difference between revisions of "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

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

\begin{align*}
&\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\\ = \ &\sqrt{\left(\sqrt{81-38}+\sqrt{81+38}\right)}\\ = \ &\sqrt{\left(\sqrt{162}\right}\\ = \ &\sqrt{\left(\sqrt{(3^4)*2\right} = \ &\boxed{2\sqrt{6} \ \mathbf{(B)}}.
\end{align*} (Error compiling LaTeX. )

(Basically, this method turns the question into a 4th root and then simplifies it. By the way, this method is much easier.) Request from the author: Can someone help fix the coding? Thx ------ SuperWill

Solution 3 (FASTER!)

We can change the insides of the square root into a perfect square and then simplify.

$\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\\ = \ &\sqrt{6-6\sqrt{2}+3}+\sqrt{6+6\sqrt{2}+3\\ = \ &\sqrt{\left{\sqrt{6}-\sqrt{3}}^2}+\sqrt{\left{\sqrt{6}+\sqrt{3}}^2}\\ = \ &\sqrt{6}-\sqrt{3}+\sqrt{6}+\sqrt{3}\\ = \ &\boxed{2\sqrt{6} \ \mathbf{(B)}}$ (Error compiling LaTeX. )

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