College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 1 - Summary, Review, and Test - Review Exercises - Page 204: 66

Answer

Solution set:$\{3,9\}$

Work Step by Step

$x^{2} -12x+27 = 0$ $x^{2} -12x =-27$ By adding the square of half the co-efficient of $x$ to both sides, we get, $x^{2} -12x+36 =-27+36$ $x^{2} -12x+36 =9$ By factoring, $(x-6)^{2}=9$ $(x-6)=±\sqrt 9$ $(x-6)=±3$ $x=6±3$ $x=6+3$ or $x=6-3$ $x=9$ or $x=3$ Solution set:$\{3,9\}$
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