Difference between revisions of "1990 AHSME Problems/Problem 1"
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== Problem== | == Problem== | ||
| − | If <math>\dfrac{x | + | If <math>\dfrac{\frac{x}{4}}{2}=\dfrac{4}{\frac{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)}\ \pm 4\qquad\text{(E)}\ \pm 8</math> | <math>\text{(A)}\ \pm\frac{1}{2}\qquad\text{(B)}\ \pm 1\qquad\text{(C)}\ \pm 2\qquad\text{(D)}\ \pm 4\qquad\text{(E)}\ \pm 8</math> | ||
Revision as of 15:55, 6 July 2014
Problem
If 
, then 
Solution
Cross-multiplying leaves
$<cmath> \begin{align*}\dfrac{x^2}{8} &= 8\\ x^2 &= 64\\ \sqrt{x^2} &= \sqrt{64}\\ x &= \pm 8\end{align*} </cmath>$ (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. 
