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

(Problem 1)
(Solution 1)
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==Solution 1==
 
==Solution 1==
 
By guessing and checking, 2 works.  
 
By guessing and checking, 2 works.  
<math>\frac{2}{x} = </math>\boxed{\textbf{(D)  }1}
+
<math>\frac{2}{x} = \boxed{\textbf{(D)  }1}</math>

Revision as of 14:24, 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 1

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