Difference between revisions of "2011 AMC 10A Problems/Problem 16"
(added a better solution) |
Erics son07 (talk | contribs) (→Solution 4- no words) |
||
(18 intermediate revisions by 12 users not shown) | |||
Line 4: | Line 4: | ||
<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. | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
− | &\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\\ = \ &\sqrt{\left(\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\right)^2}\\ = \ &\sqrt{9-6\sqrt{2}+2\sqrt{(9-6\sqrt{2})(9+6\sqrt{2})}+9+6\sqrt{2}}\\ = \ &\sqrt{18+2\sqrt{(9-6\sqrt{2})(9+6\sqrt{2})}}\\ = \ &\sqrt{18+2\sqrt{9^2-(6\sqrt{2})^2}}\\ = \ &\sqrt{18+2\sqrt{81-72}}\\ = \ &\sqrt{18+2\sqrt{9}}\\ = \ &\sqrt{18+6}\\= \ &\sqrt{24}\\ = \ &\boxed{2\sqrt{6} \ \mathbf{(B)}}. | + | &\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\\\\ = \ &\sqrt{\left(\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\right)^2}\\\\ = \ &\sqrt{9-6\sqrt{2}+2\sqrt{(9-6\sqrt{2})(9+6\sqrt{2})}+9+6\sqrt{2}}\\\\ = \ &\sqrt{18+2\sqrt{(9-6\sqrt{2})(9+6\sqrt{2})}}\\\\ = \ &\sqrt{18+2\sqrt{9^2-(6\sqrt{2})^2}}\\\\ = \ &\sqrt{18+2\sqrt{81-72}}\\\\ = \ &\sqrt{18+2\sqrt{9}}\\\\ = \ &\sqrt{18+6}\\\\= \ &\sqrt{24}\\\\ = \ &\boxed{2\sqrt{6} \ \mathbf{(B)}}. |
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | == Solution 2 == | + | == Solution 2 (FASTER!) == |
− | <cmath>\ | + | We can change the insides of the square root into a perfect square and then simplify. |
− | + | ||
− | \ | + | <cmath>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</cmath> |
+ | <cmath>= \sqrt{6-6\sqrt{2}+3}+\sqrt{6+6\sqrt{2}+3}</cmath> | ||
+ | <cmath>= \sqrt{\left(\sqrt{6}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}</cmath> | ||
+ | <cmath>= \sqrt{6}-\sqrt{3}+\sqrt{6}+\sqrt{3}</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" | ||
+ | |||
+ | |||
+ | ==Solution 4- no words== | ||
+ | <cmath>x=\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</cmath> | ||
+ | <cmath>x^{2}=9-6\sqrt{2}+9+6\sqrt{2}+2\sqrt{\left(9-6\sqrt{2}\right)\left(9+6\sqrt{2}\right)}</cmath> | ||
+ | <cmath>x^{2}=18+2\sqrt{81-72}</cmath> | ||
+ | <cmath>x^{2}=18+2\sqrt{9}</cmath> | ||
+ | <cmath>x^{2}=18+6</cmath> | ||
+ | <cmath>x^{2}=24</cmath> | ||
+ | <cmath>x=\pm\sqrt{24}</cmath> | ||
+ | <cmath>x=\pm2\sqrt{6}</cmath> | ||
+ | <cmath>\boxed{\textbf{(B) } 2\sqrt{6}}</cmath> | ||
+ | |||
+ | ~JH. L | ||
− | + | ==Video Solution== | |
− | + | https://youtu.be/ow2axpUP53c | |
− | |||
− | + | ~savannahsolver | |
− | |||
− | |||
− | |||
== See Also == | == See Also == | ||
Latest revision as of 00:39, 26 June 2022
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"
Solution 4- no words
~JH. L
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.