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



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}
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