Elementary Algebra

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

Chapter 10 - Quadratic Equations - 10.4 - Solving Quadratic Equations - Which Method? - Problem Set 10.4 - Page 457: 1



Work Step by Step

Factoring the equation and setting the factors equal to zero, we obtain: $x^{2}+4x=45$ $x^{2}+4x-45=0$ $x^{2}-5x+9x-45=0$ $x(x-5)+9(x-5)=0$ $(x-5)(x+9)=0$ $(x-5)=0$ 0r $(x+9)=0$ $x=5$ or $x=-9$ Therefore, the solution set is {$-9,5$}.
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