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: 43


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

Work Step by Step

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