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 Review Problem Set - Page 274: 36



Work Step by Step

Using the rules of factoring trinomials to factor the polynomial, we obtain: $4n^{2}+10n-14=0$ $2(2n^{2}+5n-7)=0$ $2n^{2}+5n-7=0$ $2n^{2}-2n+7n-7=0$ $2n(n-1)+7(n-1)=0$ $(n-1)(2n+7)=0$ Now, we equate the two factors to zero and solve: $(n-1)(2n+7)=0$ $(n-1)=0$ or $(2n+7)=0$ $n=1$ or $n=-\frac{7}{2}$ Therefore, the solution set is {$-\frac{7}{2},1$}.
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