Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 1-6 - Cumulative Review: 3

Answer

$t = 1, 9$

Work Step by Step

$3(t-5)^{2} - 18 = 30$ $3(t-5)^{2} = 30 + 18$ $3(t-5)^{2} = 48$ $(t-5)^{2} = 16$ $t(t-5)-5(t-5) = 16$ $t^{2} - 5t - 5t + 25 = 16$ $t^{2} - 10t + 25 = 16$ $t^{2} - 10t + 25 - 16 = 0$ $t^{2} - 10t + 9 = 0$ $t^{2}-9t-t+9 = 0$ $t(t-9)-1(t-9) = 0$ $(t - 1)(t-9) = 0$ $t = 1, 9$ Check: When $t = 1$ $3(1-5)^{2} - 18 \overset{?}{=} 30$ $3(-4)^{2} - 18 \overset{?}{=} 30$ $3(16) - 18 \overset{?}{=} 30$ $48 - 18 \overset{?}{=} 30$ $30 = 30$ When $t = 9$ $3(9-5)^{2} - 18 \overset{?}{=} 30$ $3(4)^{2} - 18 \overset{?}{=} 30$ $3(16) - 18 \overset{?}{=} 30$ $48 - 18 \overset{?}{=} 30$ $30 = 30$
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