1978 AHSME Problems/Problem 1: Difference between revisions
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..." |
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\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> | ||
==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 
, then 
equals
Solution 1
By guessing and checking, 2 works.
\boxed{\textbf{(D) }1}
