1983 AHSME Problems/Problem 1

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 $x \neq 0$ , $x=8$ . $y=16\cdot 8 = \textbf{(E)}\; 128$ .

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1983 AHSME Problems/Problem 1

Problem

Solution