Difference between revisions of "1990 AHSME Problems/Problem 1"
(Created page with '== Problem== If <math>\dfrac{x/4}{2}=\dfrac{4}{x/2}</math>, then <math>x=</math> <math>\text{(A)}\ \pm\frac{1}{2}\qquad\text{(B)}\ \pm 1\qquad\text{(C)}\ \pm 2\qquad\text{(D)}\ …') |
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So the answer is <math>\boxed{\text{(E)} \, \pm 8}</math>. | So the answer is <math>\boxed{\text{(E)} \, \pm 8}</math>. | ||
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Revision as of 13:50, 5 July 2013
Problem
If 
, then 
Solution
Cross-multiplying leaves
$<cmath> \begin{align*}\dfrac{x^2}{8} &= 8\\ x^2 &= 64\\ \sqrt{x} &= \sqrt{64}\\ x &= \pm 8\end{align*} </cmath>$ (Error compiling LaTeX. Unknown error_msg)
So the answer is 
.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
