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 (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

Video Solution

https://youtu.be/ow2axpUP53c

~savannahsolver

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