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: 48


$x=\left\{ -1,-\dfrac{1}{2} \right\}$

Work Step by Step

The factored form of the given expression, $ (2x-1)(x+2)=-3 $, is \begin{array}{} 2x^2+4x-x-2=-3 \\\\ 2x^2+3x-2+3=0 \\\\ 2x^2+3x+1=0 \\\\ (2x+1)(x+1)=0 .\end{array} Equating each factor to zero, then \begin{array}{l} 2x+1=0 \\ 2x=0-1 \\ 2x=-1 \\ x=-\dfrac{1}{2} ,\text{ OR}\\\\ x+1=0 \\ x=0-1 \\ x=-1 \end{array} Hence, $ x=\left\{ -1,-\dfrac{1}{2} \right\} $.
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