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

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== Solution ==
 
== Solution ==
  
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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}}}}$.
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Multiplying the numerator and denominator by <math>\sqrt{a^2+x^2}</math> results in  
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<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>  
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Since <math>\sqrt{a^2+x^2}=(a^2+x^2)^{\frac{1}{2}}</math>,  
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the denominator is <math>(a^2+x^2)^2\cdot (a^2+x^2)^{\frac{1}{2}} = (a^2+x^2)^{\frac{3}{2}}</math> <math>\boxed{\textbf{(D) } \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}}</math>.
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==See Also==
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{{AHSME 50p box|year=1953|num-b=6|num-a=8}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 19:40, 1 April 2017

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 $\sqrt{a^2+x^2}=(a^2+x^2)^{\frac{1}{2}}$, the denominator is $(a^2+x^2)^2\cdot (a^2+x^2)^{\frac{1}{2}} = (a^2+x^2)^{\frac{3}{2}}$ $\boxed{\textbf{(D) } \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}}$.

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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All AHSME Problems and Solutions

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