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: 48

Answer

$$ x=3.732, .268$$

Work Step by Step

The quadratic formula states: $$ x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ Solving by using the quadratic formula to approximate the solution, we find: $$ x_{1,\:2}=\frac{-\left(-4\right)\pm \sqrt{\left(-4\right)^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}\\ x=2+\sqrt{3},\:x=2-\sqrt{3} \\ x=3.732, .268$$
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