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
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

@@ Line 1: / Line 1: @@
+== Problem ==
 If the reciprocal of <math>x+1</math> is <math>x-1</math>, then <math>x</math> equals:
@@ Line 4: / Line 6: @@
 ==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
+==See Also==
+{{AHSME 40p box|year=1963|num-b=2|num-a=4}}
+[[Category:Introductory Algebra Problems]]
+{{MAA Notice}}