Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.7 - Equations - Exercise Set - Page 109: 168

Answer

$ x^{2}-2x-15=0$

Work Step by Step

$ x+3=0 \quad $ has the solution set $\{-3\}$ $(x-5)(x+3)=0\quad $ has the solution set $\{-3,5\}$ Distribute the LHS and simplify $ x^{2}-2x-15=0$
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