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...")
 
(Problem 1)
Line 8: Line 8:
 
\textbf{(D) }-1\text{ or }2\qquad  
 
\textbf{(D) }-1\text{ or }2\qquad  
 
\textbf{(E) }-1\text{ or }-2    </math>
 
\textbf{(E) }-1\text{ or }-2    </math>
 
[[1978 AHSME Problems/Problem 1|Solution]]
 
  
 
==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} = </math>\boxed{\textbf{(D)  }1}

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}