Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 6 - Section 6.6 - Solving Quadratic Equations by Factoring - Exercise Set - Page 458: 37

Answer

The solutions are 0 and $\frac{1}{2}$ and$\frac{-1}{2}$.

Work Step by Step

4$x^{3}$-x=0 x(4$x^{2}$-1)=0 x(2x-1)(2x+1)=0 x=0 or 2x-1=0 or 2x+1=0 x=0 or 2x-1+1=0+1 or 2x+1-1=0-1 x=0 or 2x=1 or 2x=-1 x=0 or x= $\frac{1}{2}$ or x=$\frac{-1}{2}$ The solutions are 0 and $\frac{1}{2}$ and$\frac{-1}{2}$. Check Let x=o 4$x^{3}$-x=0 4*$0^{3}$-0=0 0-0=0 0=0 Let x= $\frac{1}{2}$ 4$x^{3}$-x=0 4$(1/2)^{3}$-(1/2)=0 4$\frac{1}{8}$-$\frac{1}{2}$=0 $\frac{1}{2}$-$\frac{1}{2}$=0 0=0 Let x= $\frac{-1}{2}$ 4$x^{3}$-x=0 4$(-1/2)^{3}$-(-1/2)=0 4$\frac{-1}{8}$-$\frac{-1}{2}$=0 $\frac{-1}{2}$-$\frac{-1}{2}$=0 $\frac{-1}{2}$+$\frac{1}{2}$=00=0 0=0
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