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2011 AMC 10A Problems/Problem 16

Revision as of 11:05, 17 December 2020 by Coolmath 2018 (talk | contribs)

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

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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