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 5 - Polynomials and Factoring - 5.7 Solving Quadratic Equations by Factoring - 5.7 Exercise Set - Page 351: 9


$x=\left\{ -6,\dfrac{3}{2} \right\}$

Work Step by Step

Equating each factor to zero (Zero Product Principle), then the solutions to the given equation, $ (2t-3)(t+6)=0 ,$ is \begin{array}{l} 2t-3=0 \\\\ 2t=3 \\\\ t=\dfrac{3}{2} ,\\\\\text{OR}\\\\ t+6=0 \\\\ t=-6 .\end{array} Hence, $ x=\left\{ -6,\dfrac{3}{2} \right\} .$
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.