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

Latest revision as of 10:59, 13 February 2021

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 $\boxed{\textbf{(B) }1}$

1978 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 0	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 12: / Line 12: @@
 By guessing and checking, 2 works.
 <math>\frac{2}{x} = \boxed{\textbf{(B)  }1}</math>
+~awin
 ==Solution 2==
@@ Line 17: / 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>
+==See Also==
+{{AHSME box|year=1978|num-b=0|num-a=2}}
+{{MAA Notice}}

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