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Difference between revisions of "1990 AHSME Problems/Problem 1"

Problem

If $\dfrac{\frac{x}{4}}{2}=\dfrac{4}{\frac{x}{2}}$, then $x=$

$\text{(A)}\ \pm\frac{1}{2}\qquad\text{(B)}\ \pm 1\qquad\text{(C)}\ \pm 2\qquad\text{(D)}\ \pm 4\qquad\text{(E)}\ \pm 8$

Solution

Cross-multiplying leaves

\begin{align*}\dfrac{x^2}{8} &= 8\\ x^2 &= 64\\ \sqrt{x^2} &= \sqrt{64}\\ x &= \pm 8\end{align*}

So the answer is $\boxed{\text{(E)} \, \pm 8}$.