College Algebra (6th Edition)

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

Chapter 1 - Equations and Inequalities - Exercise Set 1.6 - Page 178: 7


The solutions are $y=-2$ and $y=\pm\dfrac{1}{2}$

Work Step by Step

$4y^{3}-2=y-8y^{2}$ Take all terms to the left side of the equation: $4y^{3}+8y^{2}-y-2=0$ Group the first two terms together and the last two terms together: $(4y^{3}+8y^{2})-(y+2)=0$ Take out common factor $4y^{2}$ from the first group: $4y^{2}(y+2)-(y+2)=0$ Factor out $(y+2)$: $(y+2)(4y^{2}-1)=0$ Set both factors equal to $0$ and solve each individual equation for $y$: $y+2=0$ $y=-2$ $4y^{2}-1=0$ $4y^{2}=1$ $y^{2}=\dfrac{1}{4}$ $\sqrt{y^{2}}=\pm\sqrt{\dfrac{1}{4}}$ $y=\pm\dfrac{1}{2}$ The solutions are $y=-2$ and $y=\pm\dfrac{1}{2}$
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