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 ==
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== 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 ==
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== Solution 2 (FASTER!) ==
<cmath>\begin{align*}
+
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{\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)}}.
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\end{align*}</cmath>
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<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>
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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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<cmath>x^{2}=18+6</cmath>
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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>
 +
<cmath>\boxed{\textbf{(B) } 2\sqrt{6}}</cmath>
 +
 
 +
~JH. L
  
(Basically, this method turns the question into a 4th root and then simplifies it. By the way, this method is much easier.)
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==Video Solution==
Request from the author: Can someone help fix the coding?
+
https://youtu.be/ow2axpUP53c
Thx    ------ SuperWill
 
  
== Solution 3 (FASTER!) ==
+
~savannahsolver
We can change the insides of the square root into a perfect square and then simplify.
 
  
<math>&\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}}^2}+\sqrt{\left{\sqrt{6}+\sqrt{3}}^2}\\ = \ &\sqrt{6}-\sqrt{3}+\sqrt{6}+\sqrt{3}\\ =  \ &\boxed{2\sqrt{6} \ \mathbf{(B)}}</math>
 
-----will3145
 
 
== See Also ==
 
== See Also ==
  

Latest revision as of 00:39, 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

\[x=\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}\] \[x^{2}=9-6\sqrt{2}+9+6\sqrt{2}+2\sqrt{\left(9-6\sqrt{2}\right)\left(9+6\sqrt{2}\right)}\] \[x^{2}=18+2\sqrt{81-72}\] \[x^{2}=18+2\sqrt{9}\] \[x^{2}=18+6\] \[x^{2}=24\] \[x=\pm\sqrt{24}\] \[x=\pm2\sqrt{6}\] \[\boxed{\textbf{(B) } 2\sqrt{6}}\]

~JH. L

Video Solution

https://youtu.be/ow2axpUP53c

~savannahsolver

See Also

2011 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 15
Followed by
Problem 17
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All AMC 10 Problems and Solutions

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