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: 20

Answer

$y=\pm \dfrac{4}{5}i$

Work Step by Step

Since $x^2=a$ implies $x=\pm\sqrt{a}$ (the Square Root Principle), the solutions to the given equation, $ 25y^2+16=0 ,$ are \begin{array}{l}\require{cancel} 25y^2=-16 \\\\ y^2=\dfrac{-16}{25} \\\\ y=\pm\sqrt{\dfrac{-16}{25}} \\\\ y=\pm\sqrt{-1}\cdot\sqrt{\dfrac{16}{25}} \\\\ y=\pm i\cdot\sqrt{\left( \dfrac{4}{5} \right)^2} \\\\ y=\pm i\cdot\dfrac{4}{5} \\\\ y=\pm \dfrac{4}{5}i .\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.