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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>
  
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== Solution 1 (Bash)==
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== Solution 1 (Fast)==
  
 
We find the answer by squaring, then square rooting the expression.
 
We find the answer by squaring, then square rooting the expression.
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<cmath>= \boxed{B) 2\sqrt{6}}</cmath>
 
<cmath>= \boxed{B) 2\sqrt{6}}</cmath>
  
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==Solution 3 (FASTEST)==
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==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>  
 
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>  
  
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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>
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<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>
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<cmath>\boxed{\textbf{(B) } 2\sqrt{6}}</cmath>
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 +
~JH. L
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==Solution 5 (Bash / Int. Alg.)==
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First, factor <math>\sqrt3</math> out of the whole expression: <math>\sqrt3\left(\sqrt{3-2\sqrt2}+\sqrt{3+2\sqrt2}\right)</math> Let <math>\sqrt{3-2\sqrt{2}}=a+b\sqrt2.</math> We square both sides, leaving us with <math>3-2\sqrt{2}=a^2+2ab\sqrt2+2b^2.</math> We equate coefficients, and we see that <math>3=a^2+2b^2</math> and <math>-2=2ab\implies ab=-1.</math> We can see a pair of values that works easily, <math>(a,b)=(1,-1)</math> or <math>(a,b)=(-1,1).</math> A quick sign check tells us that the valid solution is <math>(-1,1).</math>
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Similarly, for <math>\sqrt{3+2\sqrt{2}},</math> we get <math>1+\sqrt2.</math>
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When we add these two, we get <math>2\sqrt2,</math> and multiplying by <math>\sqrt3,</math> we get <math>\boxed{\text{(B)}~2\sqrt6}.</math>
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We can quickly check our answer by estimating <math>\sqrt2=1.41</math> and <math>\sqrt3=1.73</math>: <math>\sqrt3\left(\sqrt{3-2\sqrt2}+\sqrt{3+2\sqrt2}\right)</math> becomes <math>\sqrt3\left(\sqrt{3-2\cdot1.41}+\sqrt{3+2\cdot1.41}\right)=\sqrt3\left(\sqrt{3-2.82}+\sqrt{3+2.82}\right)=\sqrt3\left(\sqrt{0.18}+\sqrt{5.82}\right)\approx\sqrt3\left(\dfrac{3\sqrt2}{10}+2.4\right)\approx1.73\left(\dfrac{3\cdot1.41}{10}+2.4\right)=1.73\left(0.423+2.4\right)=1.73\left(2.823\right)\approx1.7\cdot2.8\approx4.76.</math> This should be our approximate answer. We got <math>2\sqrt6=2\sqrt2\cdot\sqrt3\approx2\cdot1.41\cdot1.73=2.82\cdot1.73\approx2.8\cdot1.7=4.76</math> - the same thing.
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~Technodoggo
  
 
==Video Solution==
 
==Video Solution==

Latest revision as of 19:43, 21 August 2023

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 (Fast)

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

Solution 5 (Bash / Int. Alg.)

First, factor $\sqrt3$ out of the whole expression: $\sqrt3\left(\sqrt{3-2\sqrt2}+\sqrt{3+2\sqrt2}\right)$ Let $\sqrt{3-2\sqrt{2}}=a+b\sqrt2.$ We square both sides, leaving us with $3-2\sqrt{2}=a^2+2ab\sqrt2+2b^2.$ We equate coefficients, and we see that $3=a^2+2b^2$ and $-2=2ab\implies ab=-1.$ We can see a pair of values that works easily, $(a,b)=(1,-1)$ or $(a,b)=(-1,1).$ A quick sign check tells us that the valid solution is $(-1,1).$ Similarly, for $\sqrt{3+2\sqrt{2}},$ we get $1+\sqrt2.$ When we add these two, we get $2\sqrt2,$ and multiplying by $\sqrt3,$ we get $\boxed{\text{(B)}~2\sqrt6}.$

We can quickly check our answer by estimating $\sqrt2=1.41$ and $\sqrt3=1.73$: $\sqrt3\left(\sqrt{3-2\sqrt2}+\sqrt{3+2\sqrt2}\right)$ becomes $\sqrt3\left(\sqrt{3-2\cdot1.41}+\sqrt{3+2\cdot1.41}\right)=\sqrt3\left(\sqrt{3-2.82}+\sqrt{3+2.82}\right)=\sqrt3\left(\sqrt{0.18}+\sqrt{5.82}\right)\approx\sqrt3\left(\dfrac{3\sqrt2}{10}+2.4\right)\approx1.73\left(\dfrac{3\cdot1.41}{10}+2.4\right)=1.73\left(0.423+2.4\right)=1.73\left(2.823\right)\approx1.7\cdot2.8\approx4.76.$ This should be our approximate answer. We got $2\sqrt6=2\sqrt2\cdot\sqrt3\approx2\cdot1.41\cdot1.73=2.82\cdot1.73\approx2.8\cdot1.7=4.76$ - the same thing. ~Technodoggo

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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