Difference between revisions of "1983 AHSME Problems/Problem 1"
(Created page with "==Problem== If <math>x \neq 0, \frac x{2} = y^2</math> and <math>\frac{x}{4} = 4y</math>, then <math>x</math> equals <math>\textbf{(A)}\ 8\qquad \textbf{(B)}\ 16\qquad \textbf...") |
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==Problem== | ==Problem== | ||
If <math>x \neq 0, \frac x{2} = y^2</math> and <math>\frac{x}{4} = 4y</math>, then <math>x</math> equals | If <math>x \neq 0, \frac x{2} = y^2</math> and <math>\frac{x}{4} = 4y</math>, then <math>x</math> equals | ||
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<math>\textbf{(A)}\ 8\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 32\qquad \textbf{(D)}\ 64\qquad \textbf{(E)}\ 128</math> | <math>\textbf{(A)}\ 8\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 32\qquad \textbf{(D)}\ 64\qquad \textbf{(E)}\ 128</math> | ||
==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>. | 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>. | ||
Revision as of 14:23, 27 June 2015
Problem
If 
and 
, then 
equals
Solution
From , we get 
. Plugging in the other equation, 
, so 
. Factoring, we get 
, so the solutions are 
and 
. Since 
, the answer is 
.
