2011 AMC 10A Problems/Problem 16
Problem 16
Which of the following is equal to 
?
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)}}.
== Solution 2 ==
&\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} = \ &\boxed{2\sqrt{6} \ \mathbf{(B)}}.
\end{align*} (Error compiling LaTeX. Unknown error_msg)
See Also
| 2011 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
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