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: 61


The solution set is {$-2+5i, -2-5i$}.

Work Step by Step

$h(x) = 5(x+2)^2$, $h(x) = -125$ For $h(x) = 5(x+2)^2$, substitute $h(x)$ with $-125$. $-125 = 5(x+2)^2$ Divide the equation by $5$. Thus, it becomes: $-25 = (x+2)^2$ Use the Square Root Property $u^2 = d$, then $u = \sqrt d$ or $u = - \sqrt d$. $x+2= \sqrt{-25}$ or $x+2= -\sqrt{-25}$ $x= -2+\sqrt{-25}$ or $x= -2-\sqrt{-25}$ $x= -2+\sqrt{25}\sqrt {-1}$ or $x= -2-\sqrt{25}\sqrt {-1}$ $x = -2+5i$ or $x=-2-5i$. The solution set is {$-2+5i, -2-5i$}.
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