Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 5 - Section 5.8 - Solving Equations by Factoring and Problem Solving - Exercise Set - Page 322: 12

Answer

{-1/2, 2}

Work Step by Step

Reorder the expression n(2n - 3) = 2 to 2$n^{2}$ - 3n - 2 = 0 1. Factor: 2$n^{2}$ - 3n - 2 = (2n + 1)(n - 2) 2. Apply the zero factor property to the factored expression: a) 2n + 1 = 0 b) n - 2 = 0 3. Solve each linear equation: a) n = -1/2 b) n = 2 The solutions are {-1/2, 2}.
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