College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 4 - Section 4.5 - Inequalities involving Quadratic Functions - 4.5 Assess Your Understanding - Page 314: 21

Answer

The inequality is valid for values less than -2/3 and more than 1.5 (not including them) i.e. $(-\infty,-\frac{2}{3})\cap (1.5,\infty)$

Work Step by Step

First, we are going to set the right side to zero and factor to find the x-intercepts: $6(x^2-1)=5x$ $6x^2-6-5x=5x-5x$ $6x^2-5x-6=0$ $(3x+2)(2x-3)=0$ $x_1=-2/3$ $x_2=1.5$ These are the critical points. We are going to take three values: one less than -2/3, one between -2/3 and 1.5, and one more than 1.5 to test in the original equation and check if the inequality is true or not: First test with a value less than -2/3: $6((-1)^2-1)>5(-1)$ $6(1-1)>-1$ $6(0)>-1$ $0>-1 \rightarrow \text{ TRUE}$ Second test with a value between -2/3 and 1.5: $6(0^2-1)>5(0)$ $6(0-1)>0$ $6(-1)>0$ $-6>0 \rightarrow \text{ FALSE}$ Third test with a value more than 1.5: $6(2^2-1)>5(2)$ $6(4-1)>10$ $6(3)>10$ $18>0 \rightarrow \text{ TRUE}$ These tests show that the inequality $6(x^2-1)>5x$ is valid for values less than -2/3 and more than 1.5 (not including them) i.e. $(-\infty,-\frac{2}{3})\cap (1.5,\infty)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.