Intermediate Algebra (12th Edition)

Published by Pearson
ISBN 10: 0321969359
ISBN 13: 978-0-32196-935-4

Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - 8.1 Exercises - Page 512: 75

Answer

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

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the nonreal complex solutions of the given equation, $ x^2=-12 ,$ take the square root of both sides (Square Root Property). Then use the properties of radicals and use $i=\sqrt{-1}.$ Finally, simplify the radical. $\bf{\text{Solution Details:}}$ Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} x=\pm\sqrt{-12} .\end{array} Using the Product Rule of radicals which is given by $\sqrt[m]{x}\cdot\sqrt[m]{y}=\sqrt[m]{xy},$ the equation above is equivalent to\begin{array}{l}\require{cancel} x=\pm\sqrt{-1}\cdot\sqrt{12} .\end{array} Using $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} x=\pm i\sqrt{12} .\end{array} Writing the radicand as an expression that contains a factor that is a perfect power of the given index and then extracting the root of that factor, the equation above is equivalent to \begin{array}{l}\require{cancel} x=\pm i\sqrt{4\cdot3} \\\\ x=\pm i\sqrt{(2)^2\cdot3} \\\\ x=\pm i(2)\sqrt{3} \\\\ x=\pm 2i\sqrt{3} .\end{array} Hence, $ x=\left\{ -2i\sqrt{3},2i\sqrt{3} \right\} .$
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