Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 11 - Cumulative Review - Page 670: 15

Answer

$x=\{2,7\}$

Work Step by Step

Let $z=(x-3)$. Then the given expression, $(x-3)^2-3(x-3)-4=0$ is equivalent to \begin{array}{l}\require{cancel} z^2-3z-4=0\\ (z-4)(z+1)=0\\ z=\{-1,4\} .\end{array} Using $z=-1$, then, \begin{array}{l} x-3=-1\\ x=2 .\end{array} Using $z=4$, then, \begin{array}{l} x-3=4\\ x=7 .\end{array} Hence, $ x=\{2,7\} .$
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