Difference between revisions of "2011 AMC 10A Problems/Problem 16"
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== Solution 2 == | == Solution 2 == | ||
| − | < | + | <math>\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)}}. | &\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*}</ | + | \end{align*}</math> |
(Basically this method turns the question into a 4th root, and then simplifies it. By the way, this method is much easier.) | (Basically this method turns the question into a 4th root, and then simplifies it. By the way, this method is much easier.) | ||
Revision as of 17:00, 8 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
$\begin{align*} &\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\\ = \ &\sqrt{\left(\sqrt{81-38}+\sqrt{81+38}\right)}\\ = \ &\sqrt{\left(\sqrt{162}\right}\\ = \ &\sqrt{\left(\sqrt{(3^4)*2\right} = \ &\boxed{2\sqrt{6} \ \mathbf{(B)}}. \end{align*}$ (Error compiling LaTeX. Unknown error_msg)
(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 fix the coding, please? Thx. ------ SuperWill
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
