Difference between revisions of "2011 AMC 10A Problems/Problem 16"
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<math>\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6</math> | <math>\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6</math> | ||
− | == Solution 1 == | + | == Solution 1 (Bash)== |
We find the answer by squaring, then square rooting the expression. | We find the answer by squaring, then square rooting the expression. | ||
Line 21: | Line 21: | ||
<cmath>= \boxed{B) 2\sqrt{6}}</cmath> | <cmath>= \boxed{B) 2\sqrt{6}}</cmath> | ||
− | ---- | + | ==Solution 3 (FASTEST)== |
+ | 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>\boxed{B) 2\sqrt{6}}</math>. | |
− | < | + | |
− | + | ~coolmath_2018 | |
− | \ | + | |
+ | Note: This is basically just Solution 2 except you "do a little experimentation" | ||
− | + | ==Video Solution== | |
− | + | https://youtu.be/ow2axpUP53c | |
− | |||
− | + | ~savannahsolver | |
== See Also == | == See Also == |
Revision as of 20:38, 16 April 2021
Contents
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.
Solution 3 (FASTEST)
Square roots remind us of squares. So lets try to make . Doing a little experimentation we find that Similarly since we know that
We want to find . Using what we found above we know This is nothing but .
~coolmath_2018
Note: This is basically just Solution 2 except you "do a little experimentation"
Video Solution
~savannahsolver
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.