Elementary Algebra

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

Chapter 11 - Additional Topics - 11.5 - Quadratic Equations: Complex Solutions - Problem Set 11.5 - Page 496: 12



Work Step by Step

Using the rules of factoring trinomials to factor, we obtain: $a^{2}+2a-35=0$ $a^{2}+7a-5a-35=0$ $a(a+7)-5(a+7)=0$ $(a+7)(a-5)=0$ $(a+7)=0$ or $(a-5)=0$ $a=-7$ or $a=5$ Therefore, the solution set is {$-7,5$}.
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