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: 7

Answer

$h = 0, -2±2\sqrt {3}$

Work Step by Step

$4h^{3} + 16h^{2} - 32h = 0$ $4h(h^{2} + 4h - 8) = 0$ $h = \frac{-(4)±\sqrt {(4)^{2}-4(1)(-8)}}{2(1)}$ $h = \frac{-4±\sqrt {16-4(1)(-8)}}{2(1)}$ $h = \frac{-4±\sqrt {16+32}}{2}$ $h = \frac{-4±\sqrt {48}}{2}$ $h = \frac{-4±2\sqrt {12}}{2}$ $h = \frac{-2±\sqrt {12}}{1}$ $h = 0, -2±\sqrt {12}$ $h = 0, -2±2\sqrt {3}$ Check: When $h = 0$ $4(0)^{3} + 16(0)^{2} - 32(0) \overset{?}{=} 0$ $0 + 0 - 0 \overset{?}{=}0$ $0 = 0$ When $h = -2+2\sqrt {3}$ $4(-2+2\sqrt {3})^{3} + 16(-2+2\sqrt {3})^{2} - 32(-2+2\sqrt {3}) \overset{?}{=} 0$ $4(3.1384...) + 16(2.14359...) - (46.851...)\overset{?}{=} 0$ $(46.851...) - (46.851...)\overset{?}{=} 0$ $0 = 0$ When $h = -2-2\sqrt {3}$ $4(-2-2\sqrt {3})^{3} + 16(-2-2\sqrt {3})^{2} - 32(-2-2\sqrt {3}) \overset{?}{=} 0$ $4(-163.1384...) + 16(29.856...) - (-174.8512...)\overset{?}{=} 0$ $(-652.5537...) + (477.702...) + (174.8512...)\overset{?}{=} 0$ $(-652.5537...) + (-652.5537...) \overset{?}{=} 0$ $0 = 0$
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