Difference between revisions of "2011 AMC 10A Problems/Problem 16"
Thedrummer (talk | contribs) (Created page with '==Problem 16== Which of the following is equal to <math>\sqrt{9-6\sqrt{2}}+\sqrt{9+6\sqrt{2}}</math>? <math>\text{(A)}\,3\sqrt2 \qquad\text{(B)}\,2\sqrt6 \qquad\text{(C)}\,\frac…') |
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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 == | ||
+ | |||
+ | We find the answer by squaring, then square rooting the expression. | ||
+ | |||
+ | <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)}}. | ||
+ | \end{align*}</cmath> | ||
+ | |||
+ | == 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> | ||
+ | |||
+ | |||
+ | == See Also == | ||
+ | |||
+ | |||
+ | {{AMC10 box|year=2011|ab=A|num-b=15|num-a=17}} | ||
+ | {{MAA Notice}} |
Revision as of 18:23, 14 January 2020
Problem 16
Which of the following is equal to ?
Solution 1
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.
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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