Difference between revisions of "2011 AMC 10A Problems/Problem 16"
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Square roots remind us of squares. So lets try to make <math>9 - 6\sqrt{2} = (a-b)^2</math>. Doing a little experimentation we find that <cmath>9 - 6\sqrt{2} = (\sqrt{6} - \sqrt{3})^2.</cmath> Similarly since <math>9 + 6\sqrt{2} = (a+b)^2</math> we know that <cmath>9 + 6\sqrt{2} = (\sqrt{6} + \sqrt{3})^2.</cmath> | Square roots remind us of squares. So lets try to make <math>9 - 6\sqrt{2} = (a-b)^2</math>. Doing a little experimentation we find that <cmath>9 - 6\sqrt{2} = (\sqrt{6} - \sqrt{3})^2.</cmath> Similarly since <math>9 + 6\sqrt{2} = (a+b)^2</math> we know that <cmath>9 + 6\sqrt{2} = (\sqrt{6} + \sqrt{3})^2.</cmath> | ||
| − | We want to find <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>. Using what we found above we know <cmath>\sqrt{9 -6\sqrt{2}}+\sqrt{9+6\sqrt{2}} = (\sqrt{6} - \sqrt{3}) + (\sqrt{6} + \sqrt{3}) = 2\sqrt{6}.</cmath> This is nothing but <math>\ | + | We want to find <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>. Using what we found above we know <cmath>\sqrt{9 -6\sqrt{2}}+\sqrt{9+6\sqrt{2}} = (\sqrt{6} - \sqrt{3}) + (\sqrt{6} + \sqrt{3}) = 2\sqrt{6}.</cmath> This is nothing but <math>\boxed{B) 2\sqrt{6}}</math>. |
| + | |||
| + | ~coolmath_2018 | ||
== See Also == | == See Also == | ||
Revision as of 11:05, 17 December 2020
Problem 16
Which of the following is equal to 
?
Solution 1 (Bash)
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)
Solution 3 (FASTEST)
Square roots remind us of squares. So lets try to make 
. Doing a little experimentation we find that ![\[9 - 6\sqrt{2} = (\sqrt{6} - \sqrt{3})^2.\]](http://latex.artofproblemsolving.com/8/8/6/886875a4c8cd0a932088de00e10bf5c3e679a98c.png)
Similarly since 
we know that ![\[9 + 6\sqrt{2} = (\sqrt{6} + \sqrt{3})^2.\]](http://latex.artofproblemsolving.com/4/6/c/46c6d8b6c258732f2c0e078a5defca6da5cba5e9.png)
We want to find . Using what we found above we know ![\[\sqrt{9 -6\sqrt{2}}+\sqrt{9+6\sqrt{2}} = (\sqrt{6} - \sqrt{3}) + (\sqrt{6} + \sqrt{3}) = 2\sqrt{6}.\]](http://latex.artofproblemsolving.com/f/6/5/f6542b60ed41bc9b34530067d0f99c588b2f9cbd.png)
This is nothing but 
.
~coolmath_2018
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. 
