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

## 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 $\boxed{\mathrm{(D)}\text{ } 3+\sqrt6 }.$