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: 5


The solutions are $x=\dfrac{3}{2}$ and $x=\pm\dfrac{1}{2}$

Work Step by Step

$2x-3=8x^{3}-12x^{2}$ Take all terms to the right side of the equation: $0=8x^{3}-12x^{2}-2x+3$ Rearrange: $8x^{3}-12x^{2}-2x+3=0$ Group the first two terms together and the last two terms together: $(8x^{3}-12x^{2})-(2x-3)=0$ Take out common factor $4x^{2}$ from the first group: $4x^{2}(2x-3)-(2x-3)=0$ Factor out $(2x-3)$: $(2x-3)(4x^{2}-1)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $2x-3=0$ $2x=3$ $x=\dfrac{3}{2}$ $4x^{2}-1=0$ $4x^{2}=1$ $x^{2}=\dfrac{1}{4}$ $\sqrt{x^{2}}=\pm\sqrt{\dfrac{1}{4}}$ $x=\pm\dfrac{1}{2}$ The solutions are $x=\dfrac{3}{2}$ and $x=\pm\dfrac{1}{2}$
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