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

(Created page with "If the reciprocal of <math>x+1</math> is <math>x-1</math>, then <math>x</math> equals: <math>\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ ...")
(No difference)

Revision as of 16:50, 1 August 2015

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

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

Solution

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

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

~mathsolver101

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