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: 11

Answer

$y=\pm\dfrac{7}{2}$

Work Step by Step

Using the properties of equality, the given equation, $ 4y^2=49 ,$ is equivalent to \begin{array}{l}\require{cancel} y^2=\dfrac{49}{4} .\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{49}{4}} \\\\ y=\pm\sqrt{\left(\dfrac{7}{2}\right)^2} \\\\ y=\pm\dfrac{7}{2} .\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.