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 706: 13

Answer

$x=\pm\dfrac{7}{3}$

Work Step by Step

Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), the solutions to the given equation, $ 9x^2-49=0 ,$ are \begin{array}{l}\require{cancel} 9x^2=49 \\\\ x^2=\dfrac{49}{9} \\\\ x=\pm\sqrt{\dfrac{49}{9}} \\\\ x=\pm\sqrt{ \left( \dfrac{7}{3} \right)^2} \\\\ x=\pm\dfrac{7}{3} .\end{array}
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.