College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 3 - Mid-Chapter Check Point - Page 390: 16

Answer

$x = -\frac{1}{2}, \frac{2}{3}, \frac{7}{2}$

Work Step by Step

Since the equation is already factorized, we can simply take each factor and equal it to zero as such: $(2x+1) = 0$, $(3x-2)^{3} = 0$ and $(2x-7)=0$ where, by solving each one individually, we get: (a)$$2x + 1 = 0$$ $$\frac{2x}{2} = -\frac{1}{2}$$ $$x=-\frac{1}{2}$$ (b)$$\sqrt[3] {(3x - 2)^{3}} = \sqrt[3] 0$$ $$3x - 2 = 0$$ $$\frac{3x}{3} = \frac{2}{3}$$ $$x = \frac{2}{3}$$ (c)$$2x - 7 = 0$$ $$\frac{2x}{2} = \frac{7}{2}$$ $$x = \frac{7}{2}$$
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