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Difference between revisions of "1954 AHSME Problems/Problem 14"

(Created page with "== Problem 14== When simplified <math>\sqrt{1+ \left (\frac{x^4-1}{2x^2} \right )^2}</math> equals: <math>\textbf{(A)}\ \frac{x^4+2x^2-1}{2x^2} \qquad \textbf{(B)}\ \frac{x...")
 
(→‎Solution)
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== Solution ==
 
== Solution ==
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<math>\sqrt{\frac{4x^4}{4x^4}+\frac{(x^4-1)^2}{4x^4}}\implies\sqrt{\frac{x^8-2x^4+1+4x^4}{4x^4}}\implies \sqrt{\frac{(x^4+1)^2}{2x^2}}\implies \frac{x^4+1}{2x^2}\implies\frac{x^2}{2}+\frac{1}{2x^2}</math>, <math>\fbox{E}</math>
+
<math>\sqrt{\frac{4x^4}{4x^4}+\frac{(x^4-1)^2}{4x^4}}\implies\sqrt{\frac{x^8-2x^4+1+4x^4}{4x^4}}\implies \sqrt{\frac{(x^4+1)^2}{(2x^2)^2}}\implies \frac{x^4+1}{2x^2}\implies\frac{x^2}{2}+\frac{1}{2x^2}</math>, <math>\fbox{E}</math>

Revision as of 13:14, 6 June 2016

Problem 14

When simplified $\sqrt{1+ \left (\frac{x^4-1}{2x^2} \right )^2}$ equals:

$\textbf{(A)}\ \frac{x^4+2x^2-1}{2x^2} \qquad \textbf{(B)}\ \frac{x^4-1}{2x^2} \qquad \textbf{(C)}\ \frac{\sqrt{x^2+1}}{2}\\ \textbf{(D)}\ \frac{x^2}{\sqrt{2}}\qquad\textbf{(E)}\ \frac{x^2}{2}+\frac{1}{2x^2}$

Solution

$\sqrt{\frac{4x^4}{4x^4}+\frac{(x^4-1)^2}{4x^4}}\implies\sqrt{\frac{x^8-2x^4+1+4x^4}{4x^4}}\implies \sqrt{\frac{(x^4+1)^2}{(2x^2)^2}}\implies \frac{x^4+1}{2x^2}\implies\frac{x^2}{2}+\frac{1}{2x^2}$, $\fbox{E}$

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