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: 17



Work Step by Step

Solving using the quadratic formula, we find: $$ \frac{1}{x^2}x^2-3x^2=\frac{8}{x}x^2\\ 1-3x^2=8x \\ -3x^2-8x+1=0\\ x_{1,\:2}=\frac{-\left(-8\right)\pm \sqrt{\left(-8\right)^2-4\left(-3\right)1}}{2\left(-3\right)}\\ x=-\frac{4+\sqrt{19}}{3},\:x=\frac{\sqrt{19}-4}{3}$$
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