# 1983 AHSME Problems/Problem 1

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## Problem

If $x \neq 0, \frac x{2} = y^2$ and $\frac{x}{4} = 4y$, then $x$ equals

$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 32\qquad \textbf{(D)}\ 64\qquad \textbf{(E)}\ 128$

## Solution

From $\frac{x}{4} = 4y$, we get $x=16y$. Plugging in the other equation, $\frac{16y}{2} = y^2$, so $y^2-8y=0$. Factoring, we get $y(y-8)=0$, so the solutions are $0$ and $8$. Since $x \neq 0$, we also have $y \neq 0$, so $y=8$. Hence $x=16\cdot 8 = \boxed{\textbf{(E)}\ 128}$.