College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 1 - Section 1.4 - Radical Equations; Equations Quadratic in Form; Factorable Equations - 1.4 Assess Your Understanding: 69

Answer

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

Work Step by Step

We solve by factoring: $3x^{-2}-7x^{-1}-6=0$ $(3x^{-1}+2)(x^{-1}-3)=0$ $(3x^{-1}+2)=0$ or $(x^{-1}-3)=0$ $3x^{-1}=-2$ or $x^{-1}=3$ $x^{-1}=-\frac{2}{3}$ or $x^{-1}=3$ $x=(-\frac{2}{3})^{-1}$ or $x=3^{-1}$ $x=-\frac{3}{2}$ or $x=\frac{1}{3}$
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