Difference between revisions of "2011 AMC 10A Problems/Problem 16"
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&\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)}}. | ||
Revision as of 14:15, 26 November 2019
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 (FASTER!)
We can change the insides of the square root into a perfect square and then simplify.
![\[\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\]](http://latex.artofproblemsolving.com/e/0/2/e02e5869af070f26c6fc4da61e8250a6eec0067f.png)
![\[= \sqrt{6-6\sqrt{2}+3}+\sqrt{6+6\sqrt{2}+3}\]](http://latex.artofproblemsolving.com/6/a/f/6aff7a724692d06cbd7e8f3df5a7be238b574439.png)
![\[= \sqrt{\left(\sqrt{6}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}\]](http://latex.artofproblemsolving.com/0/5/e/05ef80b4b73b31260ace907c23f8b53d025b03a6.png)
![\[= \sqrt{6}-\sqrt{3}+\sqrt{6}+\sqrt{3}\]](http://latex.artofproblemsolving.com/d/6/f/d6f147c33ba05c8fe913961a3c7e8a1b9c12993c.png)
![\[= \boxed{B) 2\sqrt{6}}\]](http://latex.artofproblemsolving.com/0/4/b/04b0280c3b57b9f72db0801133340f04fd95d555.png)
will3145
Solution 3 (For people who have mad skillz!)
(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
Fixed
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. 
