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

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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

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.

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

will3145

Solution 3

\begin{align*}
&\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)}}.
\end{align*} (Error compiling LaTeX. Unknown error_msg)

(Basically, this method turns the question into a 4th root and then simplifies it. By the way, this method is much easier.) Request from the author: Can someone help fix the coding? Thx ------ SuperWill

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 12: / Line 12: @@
 \end{align*}</cmath>
-== Solution 2 ==
+== Solution 2 (FASTER!) ==
+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>
+-----will3145
+== Solution 3 ==
 <cmath>\begin{align*}
 &\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)}}.
@@ Line 21: / Line 32: @@
 Thx    ------ SuperWill
-== Solution 3 (FASTER!) ==
-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>
------will3145
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