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 714: 62



Work Step by Step

To find the x-intercepts, we set the function equal to zero. Doing this, we find: $$ 0\cdot \:2\left(x-2\right)=\frac{4x-2}{x-2}\cdot \:2\left(x-2\right)+\frac{x+4}{2}\cdot \:2\left(x-2\right) \\ 0=2\left(4x-2\right)+\left(x+4\right)\left(x-2\right)\\ x^2+10x-12=0\\ x_{1,\:2}=\frac{-10\pm \sqrt{10^2-4\cdot \:1\left(-12\right)}}{2\cdot \:1}\\ x=-5+\sqrt{37},\:x=-5-\sqrt{37}$$
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