Difference between revisions of "2011 AMC 10A Problems/Problem 16"
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== Solution 2 == | == Solution 2 == | ||
<cmath>\begin{align*} | <cmath>\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) | + | &\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)} = \ &\boxed{2\sqrt{6} \ \mathbf{(B)}}. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
Revision as of 00:05, 4 February 2018
Contents
Problem 16
Which of the following is equal to 
?
Solution 1
We find the answer by squaring, then square rooting the expression.
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)} = \ &\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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
