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Difference between revisions of "2011 AMC 10A Problems/Problem 16"

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Note: This is basically just Solution 2 except you "do a little experimentation"
 
Note: This is basically just Solution 2 except you "do a little experimentation"
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==Solution 4- no words==
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<cmath>x=\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</cmath>
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<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>
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<cmath>x^{2}=18+2\sqrt{9}</cmath>
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<math></math>x^{2}=18+6$
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<cmath>x^{2}=24</cmath>
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<cmath>x=\pm\sqrt{24}</cmath>
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<cmath>x=\pm2\sqrt{6}</cmath>
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<cmath>\boxed{\textbf{(B) } 2\sqrt{6}}</cmath>
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 +
~JH. L
  
 
==Video Solution==
 
==Video Solution==

Revision as of 00:38, 26 June 2022

Problem 16

Which of the following is equal to $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$?

$\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac{7\sqrt2}{2} \qquad\text{(D)}\,3\sqrt3 \qquad\text{(E)}\,6$

Solution 1 (Bash)

We find the answer by squaring, then square rooting the expression.

\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)}}. \end{align*}

Solution 2 (FASTER!)

We can change the insides of the square root into a perfect square and then simplify.

\[\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\] \[= \sqrt{6-6\sqrt{2}+3}+\sqrt{6+6\sqrt{2}+3}\] \[= \sqrt{\left(\sqrt{6}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{3}\right)^2}\] \[= \sqrt{6}-\sqrt{3}+\sqrt{6}+\sqrt{3}\] \[= \boxed{B) 2\sqrt{6}}\]

Solution 3 (FASTEST)

Square roots remind us of squares. So lets try to make $9 - 6\sqrt{2} = (a-b)^2$. Doing a little experimentation we find that \[9 - 6\sqrt{2} = (\sqrt{6} - \sqrt{3})^2.\] Similarly since $9 + 6\sqrt{2} = (a+b)^2$ we know that \[9 + 6\sqrt{2} = (\sqrt{6} + \sqrt{3})^2.\]

We want to find $\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}$. Using what we found above we know \[\sqrt{9 -6\sqrt{2}}+\sqrt{9+6\sqrt{2}} = (\sqrt{6} - \sqrt{3}) + (\sqrt{6} + \sqrt{3}) = 2\sqrt{6}.\] This is nothing but $\boxed{B) 2\sqrt{6}}$.

~coolmath_2018

Note: This is basically just Solution 2 except you "do a little experimentation"


Solution 4- no words

\[x=\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\] \[x^{2}=9-6\sqrt{2}+9+6\sqrt{2}+2\sqrt{\left(9-6\sqrt{2}\right)\left(9+6\sqrt{2}\right)}\] \[x^{2}=18+2\sqrt{81-72}\] \[x^{2}=18+2\sqrt{9}\] $$ (Error compiling LaTeX. Unknown error_msg)x^{2}=18+6$ \[x^{2}=24\] \[x=\pm\sqrt{24}\] \[x=\pm2\sqrt{6}\] \[\boxed{\textbf{(B) } 2\sqrt{6}}\]

~JH. L

Video Solution

https://youtu.be/ow2axpUP53c

~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

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