# 1953 AHSME Problems/Problem 7

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## Problem

The fraction $\frac{\sqrt{a^2+x^2}-\frac{x^2-a^2}{\sqrt{a^2+x^2}}}{a^2+x^2}$ reduces to:

$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{2a^2}{a^2+x^2} \qquad \textbf{(C)}\ \frac{2x^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad \textbf{(D)}\ \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}\qquad \textbf{(E)}\ \frac{2x^2}{a^2+x^2}$

## Solution

Multiplying the numerator and denominator by $\sqrt{a^2+x^2}$ results in $$\frac{a^2+x^2-x^2+a^2}{(a^2+x^2)(\sqrt{a^2+x^2)}}=\frac{2a^2}{(a^2+x^2)(\sqrt{a^2+x^2)}}.$$ Since $\sqrt{a^2+x^2}=(a^2+x^2)^{\frac{1}{2}}$, if we call $a^2+x^2=p$, the denominator is really just $p^1\cdot{p^{\frac{1}{2}}}=p^{\frac{3}{2}}=(a^2+x^2)^{\frac{3}{2}}$. So, the fraction is just $\boxed{\textbf{(D) } \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}}$.