Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Section 9.1 - Solving Quadratic Equations by the Square Root Property - Exercise Set - Page 631: 12



Work Step by Step

Using the properties of equality, the given equation, $ 16y^2=25 ,$ is equivalent to \begin{array}{l}\require{cancel} y^2=\dfrac{25}{16} .\end{array} Taking the square root of both sides, then the solutions of the given equation are \begin{array}{l}\require{cancel} y=\pm\sqrt{\dfrac{25}{16}} \\\\ y=\pm\sqrt{\left(\dfrac{5}{4}\right)^2} \\\\ y=\pm\dfrac{5}{4} .\end{array}
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