# Difference between revisions of "1953 AHSME Problems/Problem 7"

(Created page with "== Problem == The fraction <math>\frac{\sqrt{a^2+x^2}-\frac{x^2-a^2}{\sqrt{a^2+x^2}}}{a^2+x^2}</math> reduces to: <math>\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{2a^2}{a^2...") |
Rastyrocky (talk | contribs) (→Solution) |
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== Solution == | == Solution == | ||

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

## Revision as of 14:11, 29 October 2016

## Problem

The fraction reduces to:

## Solution

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