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: 16

Answer

$x=\left\{ -\dfrac{1}{10},\dfrac{1}{27} \right\}$

Work Step by Step

Equating each factor to zero (Zero Product Principle), then the solutions to the equation, $ \left( \dfrac{1}{9}-3x \right)\left( \dfrac{1}{5}+2x \right)=0 ,$ is \begin{array}{l}\require{cancel} \dfrac{1}{9}-3x=0 \\\\ -3x=-\dfrac{1}{9} \\\\ x=\dfrac{-\dfrac{1}{9}}{-3} \\\\ x=\dfrac{1}{27} ,\\\\\text{OR}\\\\ \dfrac{1}{5}+2x=0 \\\\ 2x=-\dfrac{1}{5} \\\\ x=-\dfrac{\dfrac{1}{5}}{2} \\\\ x=-\dfrac{1}{10} .\end{array} Hence, $ x=\left\{ -\dfrac{1}{10},\dfrac{1}{27} \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.