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 732: 32

Answer

$m=\displaystyle \frac{2}{3}$ or $m=-\displaystyle \frac{1}{5}$

Work Step by Step

$ 2m^{-2}+7m^{-1}-15=0\qquad$...substitute $m^{-1}$ for $u$ so that $u^{2}=m^{-2}$ $ 2u^{2}+7u-15=0\qquad$... solve with the Quadratic formula. $u=\displaystyle \frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ $u=\displaystyle \frac{-7\pm\sqrt{49+120}}{4}$ $u=\displaystyle \frac{-7\pm\sqrt{169}}{4}$ $u=\displaystyle \frac{-7\pm 13}{4}$ $u=\displaystyle \frac{-7+13}{4}=\frac{6}{4}=\frac{3}{2}$ or $u=\displaystyle \frac{-7-13}{4}=\frac{-20}{4}=-5$ Bring back $m^{-1}=u$. $m^{-1}=\displaystyle \frac{3}{2}$ or $ m^{-1}=-5\qquad$...recall that $m^{-1}=\displaystyle \frac{1}{m}$ $\displaystyle \frac{1}{m}=\frac{3}{2}$ or $\displaystyle \frac{1}{m}=-5 $ $m=\displaystyle \frac{2}{3}$ or $m=-\displaystyle \frac{1}{5}$
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