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 - Page 655: 23


$y=\left\{ -3-3i\sqrt{2},-3+3i\sqrt{2} \right\}$

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+3)^2=-18 ,$ are \begin{array}{l}\require{cancel} y+3=\pm\sqrt{-18} \\\\ y+3=\pm\sqrt{-1}\cdot\sqrt{18} \\\\ y+3=\pm i\cdot\sqrt{9\cdot2} \\\\ y+3=\pm i\cdot\sqrt{(3)^2\cdot2} \\\\ y+3=\pm i\cdot3\sqrt{2} \\\\ y+3=\pm 3i\sqrt{2} \\\\ y=-3\pm 3i\sqrt{2} .\end{array} Hence, $ y=\left\{ -3-3i\sqrt{2},-3+3i\sqrt{2} \right\} .$
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