Elementary Algebra

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

Chapter 6 - Factoring, Solving Equations, and Problem Solving - Chapter 6 Test - Page 275: 11



Work Step by Step

Since $7$ is common to both the terms of the equation, we take it out as a common factor: $7x^{2}=63$ $7x^{2}-63=0$ $7(x^{2}-9)=0$ $(x^{2}-9)=0$ We simplify the expression further using the rule $a^{2}-b^{2}=(a+b)(a-b)$: $(x^{2}-9)=0$ $(x^{2}-3^{2})=0$ $(x+3)(x-3)=0$ Now, we equate all of the factors to zero to solve the equation: $(x+3)(x-3)=0$ $x+3=0$ or $(x-3)=0$ $x=-3$ or $x=3$ Therefore, the solution set is {$-3,3$}.
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