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 - 11.1 Quadratic Equations - 11.1 Exercise Set - Page 707: 72



Work Step by Step

To find the x-intercepts, we set the equation equal to zero and then complete the square. Doing this, we find: $$ 3x^2+x=5\\ x^2+\frac{x}{3}=\frac{5}{3}\\ x^2+\frac{x}{3}+\left(\frac{1}{6}\right)^2=\frac{5}{3}+\left(\frac{1}{6}\right)^2\\ \left(x+\frac{1}{6}\right)^2=\frac{61}{36}\\ x=\frac{\sqrt{61}-1}{6},\:x=\frac{-\sqrt{61}-1}{6}$$
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