Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.1 - The Square Root Property and Completing the Square - Exercise Set - Page 593: 62


The solution set is {$x=4+2i, x=4-2i$}.

Work Step by Step

$h(x) = 3(x-4)^2$, $h(x)= -12$ For $h(x) = 3(x-4)^2$, substitute $h(x)$ with $-12$. $-12 = 3(x-4)^2$ Divide the equation by $3$. Thus, it becomes: $-4 = (x-4)^2$ Use the Square Root Property $u^2 = d$, then $u = \sqrt d$ or $u = - \sqrt d$. $x-4 = \sqrt{-4}$ or $x-4 = -\sqrt{-4}$ $x = 4+\sqrt{-4}$ or $x = 4-\sqrt{-4}$ $x = 4+\sqrt4\sqrt{-1}$ or $x = 4-\sqrt4\sqrt{-1}$ $x=4+2i$ or $x=4-2i$ The solution set is {$x=4+2i, x=4-2i$}.
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