College Algebra (10th Edition)

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

Chapter 1 - Section 1.1 - Linear Equations - 1.1 Assess Your Understanding - Page 90: 46

Answer

$x=\frac{-15}{7}$

Work Step by Step

$\displaystyle \begin{array}{{>{\displaystyle}l}} We\ have:\ ( x+2)( x-3) =( x+3)^{2}\\ \\ First,\ expand\ all\ brackets:\\ \\ ( x+2)( x-3) =( x+3)^{2}\\ x^{2} -3x+2x-6=x^{2} +6x+9\\ \\ Then\ simplify:\\ x^{2} -x-6=x^{2} +6x+9\\ -6=7x+9\\ 7x=-15\\ x=\frac{-15}{7} \end{array}$
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