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 (Fast)== |
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 | ||
+ | |||
+ | ==Solution 5 (Bash / Int. Alg.)== | ||
+ | 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> | ||
+ | Similarly, for <math>\sqrt{3+2\sqrt{2}},</math> we get <math>1+\sqrt2.</math> | ||
+ | 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> | ||
+ | |||
+ | 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. | ||
+ | ~Technodoggo | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/ow2axpUP53c | ||
+ | |||
+ | ~savannahsolver | ||
== See Also == | == See Also == |
Latest revision as of 20:43, 21 August 2023
Contents
Problem 16
Which of the following is equal to ?
Solution 1 (Fast)
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
Solution 5 (Bash / Int. Alg.)
First, factor out of the whole expression:
Let
We square both sides, leaving us with
We equate coefficients, and we see that
and
We can see a pair of values that works easily,
or
A quick sign check tells us that the valid solution is
Similarly, for
we get
When we add these two, we get
and multiplying by
we get
We can quickly check our answer by estimating and
:
becomes
This should be our approximate answer. We got
- the same thing.
~Technodoggo
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.