1978 AHSME Problems/Problem 1: Difference between revisions

Revision as of 20:16, 25 February 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. $\frac{2}{x} = \boxed{\textbf{(B) }1}$ ~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 $\boxed{\textbf{(B) }1}$

@@ Line 18: / Line 18: @@
 <math>\frac{2}{x} = \boxed{\textbf{(B)  }1}</math>
 ~awin
+==Solution 3==
+Directly factoring, we get <math>(1-\frac{2}{x})^2 = 0</math>. Thus <math>\frac{2}{x}</math> must equal <math>\boxed{\textbf{(B)  }1}</math>

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