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

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==Problem==
 
==Problem==
  
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The expression <math>\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}</math> is equal to:  
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The expression <math>\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}}</math> is equal to:  
  
 
<math> \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 </math>
 
<math> \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 </math>
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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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{{AHSME 40p box|year=1962|before=Problem 1|num-a=3}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 22:45, 26 December 2015

Problem

The expression $\sqrt{\frac{4}{3}} - \sqrt{\frac{3}{4}}$ 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

1962 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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