Elementary Algebra

Published by Cengage Learning
ISBN 10: 1285194055
ISBN 13: 978-1-28519-405-9

Chapter 10 - Quadratic Equations - Chapter 10 Review Problem Set: 14

Answer

{$-3,9$}

Work Step by Step

Using the rules for the factoring of trinomials, we obtain: $x(x-6)=27$ $x^{2}-6x-27=0$ $x^{2}+3x-9x-27=0$ $x(x+3)-9(x+3)=0$ $(x+3)(x-9)=0$ $(x+3)=0$ and $(x-9)=0$ $x=-3$ and $x=9$ Therefore, the solution set is {$-3,9$}.
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