Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.5 Equations Reducible to Quadratic - 11.5 Exercise Set - Page 731: 29


$x=\displaystyle \frac{1}{3}$ or $x=-\displaystyle \frac{1}{2}$

Work Step by Step

$ x^{-2}-x^{-1}-6=0\qquad$...substitute $x^{-1}$ for $u$ so that $u^{2}=x^{-2}$ $ u^{2}-u-6=0\qquad$... solve with the Quadratic formula. $u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $u=\displaystyle \frac{1\pm\sqrt{1+24}}{2}$ $u=\displaystyle \frac{1\pm\sqrt{25}}{2}$ $u=\displaystyle \frac{1\pm 5}{2}$ $u=\displaystyle \frac{1+5}{2}=\frac{6}{2}=3$ or $u=\displaystyle \frac{1-5}{2}=\frac{-4}{2}=-2$ Bring back $x^{-1}=u$. $x^{-1}=3$ or $ x^{-1}=-2\qquad$...recall that $x^{-1}=\displaystyle \frac{1}{x}$ $\displaystyle \frac{1}{x}=3$ or $\displaystyle \frac{1}{x}=-2\qquad$...multiply both sides by $x$. $1=3x$ or $ 1=-2x\qquad$...divide the first expression with $3$ and the second with $-2$. $x=\displaystyle \frac{1}{3}$ or $x=-\displaystyle \frac{1}{2}$
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