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Difference between revisions of "1978 AHSME Problems/Problem 1"

(Created page with "== Problem 1 == If <math>1-\frac{4}{x}+\frac{4}{x^2}=0</math>, then <math>\frac{2}{x}</math> equals <math>\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \te...")
(No difference)

Revision as of 15:24, 20 January 2020

Problem 1

If $1-\frac{4}{x}+\frac{4}{x^2}=0$, then $\frac{2}{x}$ equals

$\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }-1\text{ or }2\qquad \textbf{(E) }-1\text{ or }-2$

Solution

Solution 1

By guessing and checking, 2 works. $\frac{2}{x} =$\boxed{\textbf{(D) }1}

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