1978 AHSME Problems/Problem 1

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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. $\frac{2}{x} = \boxed{\textbf{(B)  }1}$ ~awin

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

Solution 3

Directly factoring, we get $(1-\frac{2}{x})^2 = 0$. Thus $\frac{2}{x}$ must equal $\boxed{\textbf{(B)  }1}$

See Also

1978 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 0
Followed by
Problem 2
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