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Difference between revisions of "1962 AHSME Problems/Problem 2"

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Since we cannot simplify further, the correct answer is <math>\boxed{\textbf{(A)}\ \frac{\sqrt{3}}{6}}</math>
 
Since we cannot simplify further, the correct answer is <math>\boxed{\textbf{(A)}\ \frac{\sqrt{3}}{6}}</math>
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==See Also==
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Revision as of 21:24, 10 November 2013

Problem

The expression $\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}$ (Error compiling LaTeX. Unknown error_msg) is equal to:

$\textbf{(A)}\ \frac{\sqrt{3}}{6}\qquad\textbf{(B)}\ \frac{-\sqrt{3}}{6}\qquad\textbf{(C)}\ \frac{\sqrt{-3}}{6}\qquad\textbf{(D)}\ \frac{5\sqrt{3}}{6}\qquad\textbf{(E)}\ 1$

Solution

Simplifying $\sqrt{\dfrac{4}{3}}$ yields $\dfrac{2}{\sqrt{3}}=\dfrac{2\sqrt{3}}{3}$.


Simplifying $\sqrt{\dfrac{3}{4}}$ yields $\dfrac{\sqrt{3}}{2}$.


$\dfrac{2\sqrt{3}}{3}-\dfrac{\sqrt{3}}{2}=\dfrac{4\sqrt{3}}{6}-\dfrac{3\sqrt{3}}{6}=\dfrac{\sqrt{3}}{6}$.


Since we cannot simplify further, the correct answer is $\boxed{\textbf{(A)}\ \frac{\sqrt{3}}{6}}$

See Also

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

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