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: 61

Answer

The solutions are $x=-1,0$ and $x=3$

Work Step by Step

$4(x+1)^{1/2}-5(x+1)^{3/2}+(x+1)^{5/2}=0\hspace{0.7cm} \color{blue}{\text{Given equation}}$ $(x+1)^{1/2}\bigg(4-5(x+1)+(x+1)^2\bigg)=0$ $(x+1)^{1/2}\bigg(4-5(x+1)+(x^2+2x+1)\bigg)=0$ $(x+1)^{1/2}\bigg(x^2-3x\bigg)=0$ $(x+1)^{1/2}x(x-3)=0$ $(x+1)^{1/2}=0\hspace{0.5cm}or\hspace{0.5cm}x=0\hspace{0.5cm}or\hspace{0.5cm}x-3=0$ $x=-1\hspace{0.5cm}or\hspace{0.5cm}x=0\hspace{0.5cm}or\hspace{0.5cm}x=3$ The solutions are $x=-1,0$ and $x=3$
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