Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

Published by Pearson
ISBN 10: 0-32184-874-8
ISBN 13: 978-0-32184-874-1

Chapter 11 - Quadratic Functions and Equations - 11.2 The Quadratic Formula - 11.2 Exercise Set - Page 713: 42



Work Step by Step

Solving by plugging in the given value and then using the quadratic formula, we find: $$ 1\cdot \:x\left(x+3\right)=\frac{2}{x}x\left(x+3\right)+\frac{2}{x+3}x\left(x+3\right)\\ x\left(x+3\right)=2\left(x+3\right)+2x\\ x^2-x-6=0\\ x_{1,\:2}=\frac{-\left(-1\right)\pm \sqrt{\left(-1\right)^2-4\cdot \:1\left(-6\right)}}{2\cdot \:1}\\ x=3,\:x=-2$$
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