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: 8


The solutions are $y=2$ and $y=\pm\dfrac{2}{3}$

Work Step by Step

$9y^{3}+8=4y+18y^{2}$ Take all terms to the left side of the equation: $9y^{3}-18y^{2}-4y+8=0$ Group the first two terms together and the last two terms together: $(9y^{3}-18y^{2})-(4y-8)=0$ Take out common factor $9y^{2}$ from the first group and common factor $4$ from the second group: $9y^{2}(y-2)-4(y-2)=0$ Factor out $(y-2)$: $(y-2)(9y^{2}-4)=0$ Set both factors equal to $0$ and solve each individual equation for $y$: $y-2=0$ $y=2$ $9y^{2}-4=0$ $9y^{2}=4$ $y^{2}=\dfrac{4}{9}$ $\sqrt{y^{2}}=\pm\sqrt{\dfrac{4}{9}}$ $y=\pm\dfrac{2}{3}$ The solutions are $y=2$ and $y=\pm\dfrac{2}{3}$
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