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 457: 12


The solutions are $\frac{2}{3}$ and $\frac{-1}{5}$.

Work Step by Step

(3x-2)(5x+1)=0 3x-2=0 or 5x+1=0 3x-2+2=0+2 or 5x+1-1=0-1 3x=2 or 5x=-1 $\frac{3x}{3}$=$\frac{2}{3}$ or $\frac{5x}{5}$=$\frac{-1}{5}$ x=$\frac{2}{3}$ or x=$\frac{-1}{5}$ The solutions are $\frac{2}{3}$ and $\frac{-1}{5}$. check let x=$\frac{-1}{5}$ (3x-2)(5x+1)=0 [3(-$\frac{1}{5}$ ) -2][5 (-$\frac{1}{5}$ )+1]= [3(-$\frac{1}{5}$ ) -2] [-1+1]]=0 [3(-$\frac{1}{5}$ ) -2]0=0 0=0 let x=$\frac{2}{3}$ (3x-2)(5x+1)=0 (3$\frac{2}{3}$ -2)(5$\frac{2}{3}$ +1)=0 (3$\frac{2}{3}$ -2)(5$\frac{2}{3}$ +1 =0 (2-2)(5$\frac{2}{3}$ +1)=0 0=0
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