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: 14

Answer

The solutions are $\frac{3}{4}$ and $\frac{-1}{9}$.

Work Step by Step

(9x+1)(4x-3)=0 4x-3=0 or 9x+1=0 4x-3+3=0+3 or 9x+1-1=0-1 4x=3 or 9x=-1 $\frac{4x}{4}$=$\frac{3}{4}$ or $\frac{9x}{9}$=$\frac{-1}{9}$ x=$\frac{3}{4}$ or x=$\frac{-1}{9}$ The solutions are $\frac{3}{4}$ and $\frac{-1}{9}$. check Let x=$\frac{3}{4}$ (9x+1)(4x-3)=0 (9$\frac{3}{4}$ +1)(4$\frac{3}{4}$ -3)=0 (9$\frac{3}{4}$ +1)(3 -3)=0 (9$\frac{3}{4}$ +1)0=0 0=0 Let x= $\frac{-1}{9}$ (9x+1)(4x-3)=0 [9$\frac{(-1)}{9}$+1][4($\frac{-1}{9}$)-3]=0 (-1+1)[4($\frac{-1}{9}$)-3]=0 0)[4($\frac{-1}{9}$)-3]=0 0=0
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