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 9.4 - Imaginary Numbers as Solutions of Quadratic Equations - Exercise Set: 24

Answer

$y=\left\{ -4-4i\sqrt{3},-4+4i\sqrt{3} \right\}$

Work Step by Step

Taking the square root of both sides (the Square Root Property) and using $i=\sqrt{-1}$, the solutions to the given equation, $ (y+4)^2=-48 ,$ are \begin{array}{l}\require{cancel} y+4=\pm\sqrt{-48} \\\\ y+4=\pm\sqrt{-1}\cdot\sqrt{48} \\\\ y+4=\pm i\cdot\sqrt{16\cdot3} \\\\ y+4=\pm i\cdot\sqrt{(4)^2\cdot3} \\\\ y+4=\pm 4i\sqrt{3} \\\\ y=-4\pm 4i\sqrt{3} .\end{array} Hence, $ y=\left\{ -4-4i\sqrt{3},-4+4i\sqrt{3} \right\} .$
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.