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: 4


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

Work Step by Step

$4x^{3}-12x^{2}=9x-27$ Take all terms to the left side of the equation: $4x^{3}-12x^{2}-9x+27=0$ Group the first two terms together and the last two terms together: $(4x^{3}-12x^{2})-(9x-27)=0$ Take out common factor $4x^{2}$ from the first group and common factor $9$ from the second group: $4x^{2}(x-3)-9(x-3)=0$ Factor out $(x-3)$: $(x-3)(4x^{2}-9)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-3=0$ $x=3$ $4x^{2}-9=0$ $4x^{2}=9$ $x^{2}=\dfrac{9}{4}$ $\sqrt{x^{2}}=\pm\sqrt{\dfrac{9}{4}}$ $x=\pm\dfrac{3}{2}$ The solutions are $x=3$ and $x=\pm\dfrac{3}{2}$
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