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

(→‎Solution 1)
(→‎Solution 2)
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Multiplying each side by <math>x^2</math>, we get <math>x^2-4x+4 = 0</math>. Factoring, we get <math>(x-2)(x-2) = 0</math>. Therefore, <math>x = 2</math>.  
 
Multiplying each side by <math>x^2</math>, we get <math>x^2-4x+4 = 0</math>. Factoring, we get <math>(x-2)(x-2) = 0</math>. Therefore, <math>x = 2</math>.  
 
<math>\frac{2}{x} = \boxed{\textbf{(B)  }1}</math>
 
<math>\frac{2}{x} = \boxed{\textbf{(B)  }1}</math>
 +
~awin

Revision as of 14:26, 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. $\frac{2}{x} = \boxed{\textbf{(B) }1}$

Solution 2

Multiplying each side by $x^2$, we get $x^2-4x+4 = 0$. Factoring, we get $(x-2)(x-2) = 0$. Therefore, $x = 2$. $\frac{2}{x} = \boxed{\textbf{(B) }1}$ ~awin

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