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: 56

Answer

The solutions are $x=-\frac{1}{2}$ and $x=-\frac{1}{4}$

Work Step by Step

$\bigg(\dfrac{x+1}{x}\bigg)^2+4\bigg(\dfrac{x+1}{x}\bigg)+3=0\hspace{0.7cm} \color{blue}{\text{Given equation}}$ $W^2+4W+3=0\hspace{3.3cm} \color{blue}{\text{Let } W=\dfrac{x+1}{x}}$ $(W+1)+(W+3)=0\hspace{2.3cm} \color{blue}{\text{Factor}}$ $W+1=0\hspace{0.5cm}or\hspace{0.5cm}W+3=0\hspace{1.3cm} \color{blue}{\text{Zero-Product Property}}$ $W=-1\hspace{1.5cm}W=-3\hspace{1.99cm} \color{blue}{\text{Solve}}$ Now change $W$ back into the correspondimg values of $x$ $\dfrac{x+1}{x}=-1\hspace{1.5cm}\dfrac{x+1}{x}=-3$ $x+1=-x\hspace{1.5cm}x+1=-3x$ $-2x=1\hspace{1.5cm}4x+1=-1$ $x=-\frac{1}{2}\hspace{1.5cm}x=-\frac{1}{4}$ The solutions are $x=-\frac{1}{2}$ and $x=-\frac{1}{4}$
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