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 323: 15

Answer

{-3, 6}

Work Step by Step

1. Factor the expression. $z^{2}$/6 - z/2 - 3 = (z/3 + 1)(z/2 - 3) 2. Apply the zero factor property. a) z/3 + 1 = 0 b) z/2 - 3 = 0 3. Solve each linear equation. a) z = -3 b) z = 6 The solutions are {-3, 6}.
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