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: 14



Work Step by Step

Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), the solutions to the given equation, $ 36a^2-25=0 ,$ are \begin{array}{l}\require{cancel} 36a^2=25 \\\\ a^2=\dfrac{25}{36} \\\\ a=\pm\sqrt{\dfrac{25}{36}} \\\\ a=\pm\sqrt{ \left( \dfrac{5}{6} \right)^2} \\\\ a=\pm\dfrac{5}{6} .\end{array}
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