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: 11

Answer

The solutions are $\frac{-3}{2}$ and $\frac{5}{4}$.

Work Step by Step

(2x+3)(4x-5)=0 2x+3=0 or 4x-5=0 2x+3-3=0-3 or 4x-5+5=0+5 2x=-3 or 4x=5 $\frac{2x}{2}$=$\frac{-3}{2}$ or $\frac{4x}{4}$=$\frac{5}{4}$ x=-$\frac{3}{2}$ or x=$\frac{5}{4}$ The solutions are $\frac{-3}{2}$ and $\frac{5}{4}$. check let x=-$\frac{3}{2}$ (2x+3)(4x-5)=0 [2(-$\frac{3}{2}$ ) +3][4 (-$\frac{3}{2}$ )-5]= (-3+3) [4 (-$\frac{3}{2}$ )-5]=0 0[4 (-$\frac{3}{2}$ )-5]=0 0=0 let x=$\frac{5}{4}$ (2x+3)(4x-5)=0 (2$\frac{5}{4}$ +3)(4 $\frac{5}{4}$ -5)=0 (2$\frac{5}{4}$ +3) (5-5)=0 (2$\frac{5}{4}$ +3) 0=0 0=0
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