Difference between revisions of "1963 AHSME Problems/Problem 3"

Latest revision as of 12:55, 16 July 2024

If the reciprocal of $x+1$ is $x-1$ , then $x$ equals:

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ \pm 1\qquad \textbf{(E)}\ \text{none of these}$

We form the equation $x+1=\frac{1}{x-1}$ .

Getting rid of the fraction yields: $x^2-1=1$ $\implies$ $x^2=2$ $\implies$ $x=\pm{\sqrt{2}}=\boxed{\text{E}}$

~mathsolver101

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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All AHSME Problems and Solutions

Revision as of 20:06, 10 January 2023 (view source) Megaboy6679 (talk \| contribs) m ← Older edit		Latest revision as of 12:55, 16 July 2024 (view source) Tecilis459 (talk \| contribs) m (Add problem header)
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		+	== Problem ==
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	If the reciprocal of <math>x+1</math> is <math>x-1</math>, then <math>x</math> equals:		If the reciprocal of <math>x+1</math> is <math>x-1</math>, then <math>x</math> equals: