College Algebra (10th Edition)

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

Chapter 1 - Review Exercises - Page 145: 6


$x=-3$ or $x=2$

Work Step by Step

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