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

Revision as of 17:50, 1 August 2015

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

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

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

@@ Line 4: / Line 4: @@
 ==Solution==
-We form the equation <math>x+1=\frac{1}{x-1}</math>.  G
+We form the equation <math>x+1=\frac{1}{x-1}</math>.
-etting rid of the fraction yields: <math>x^2-1=1</math> <math>\implies</math> <math>x^2=2</math> <math>\implies</math> <math>x=\pm{\sqrt{2}}=\boxed{\text{E}}</math>
+Getting rid of the fraction yields: <math>x^2-1=1</math> <math>\implies</math> <math>x^2=2</math> <math>\implies</math> <math>x=\pm{\sqrt{2}}=\boxed{\text{E}}</math>
 ~mathsolver101