College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises - Page 139: 67

Answer

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

Work Step by Step

$\displaystyle \frac{1}{x^{3}}+\frac{4}{x^{2}}+\frac{4}{x}=0$ We multiply through by $x^3$: $1+4x+4x^{2}=0$ Now we factor and solve: $(1+2x)^{2}=0$ $1+2x=0$ $2x=-1$ $x=-\frac{1}{2}$
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