Difference between revisions of "1950 AHSME Problems/Problem 10"

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<cmath>\frac{\sqrt3 - \sqrt2}{\sqrt3}\cdot \frac{\sqrt3 + \sqrt2}{\sqrt3 + \sqrt2} = \frac{1}{3+\sqrt6}.</cmath>
 
<cmath>\frac{\sqrt3 - \sqrt2}{\sqrt3}\cdot \frac{\sqrt3 + \sqrt2}{\sqrt3 + \sqrt2} = \frac{1}{3+\sqrt6}.</cmath>
  
The denominator is <math>\boxed{\mathrm{(D)}\text{ } 3+\sqrt6 .}</math>
+
The denominator is <math>\boxed{\mathrm{(D)}\text{ } 3+\sqrt6 }.</math>
  
 
==See Also==
 
==See Also==

Latest revision as of 10:37, 20 December 2018

Problem

After rationalizing the numerator of $\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}$, the denominator in simplest form is:

$\textbf{(A)}\ \sqrt{3}(\sqrt{3}+\sqrt{2})\qquad\textbf{(B)}\ \sqrt{3}(\sqrt{3}-\sqrt{2})\qquad\textbf{(C)}\ 3-\sqrt{3}\sqrt{2}\qquad\\ \textbf{(D)}\ 3+\sqrt6\qquad\textbf{(E)}\ \text{None of these answers}$

Solution

To rationalize the numerator, multiply by the conjugate.

\[\frac{\sqrt3 - \sqrt2}{\sqrt3}\cdot \frac{\sqrt3 + \sqrt2}{\sqrt3 + \sqrt2} = \frac{1}{3+\sqrt6}.\]

The denominator is $\boxed{\mathrm{(D)}\text{ } 3+\sqrt6 }.$

See Also

1950 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
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