Difference between revisions of "1983 AHSME Problems/Problem 1"

Revision as of 14:13, 28 June 2015

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

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

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	==Solution==		==Solution==
−	From <math>\frac{x}{4} = 4y</math>, we get <math>x=16y</math>. Plugging in the other equation, <math>\frac{16y}{2} = y^2</math>, so <math>y^2-8y=0</math>. Factoring, we get <math>y(y-8)=0</math>, so the solutions are <math>0</math> and <math>8</math>. Since <math>x \neq 0</math>, the answer is <math>\textbf{(A)}\ 8</math>.