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: 78

Answer

$t=\left\{ -6-3i,-6+3i \right\}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To find the non real complex solutions of the given equation, $ (t+6)^2=-9 ,$ 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 and isolate the variable. $\bf{\text{Solution Details:}}$ Taking the square root of both sides (Square Root Property), the equation above is equivalent to \begin{array}{l}\require{cancel} t+6=\pm\sqrt{-9} .\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} t+6=\pm\sqrt{-1}\cdot\sqrt{9} .\end{array} Using $i=\sqrt{-1},$ the equation above is equivalent to \begin{array}{l}\require{cancel} t+6=\pm i\sqrt{9} .\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} t+6=\pm i\sqrt{(3)^2} \\\\ t+6=\pm i(3) \\\\ t+6=\pm 3i .\end{array} Using the properties of equality to isolate the variable, the equation above is equivalent to \begin{array}{l}\require{cancel} t=-6\pm 3i .\end{array} Hence, $ t=\left\{ -6-3i,-6+3i \right\} .$
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