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: 13

Answer

The solutions are $\frac{7}{2}$ and $\frac{-2}{7}$

Work Step by Step

(2x-7)(7x+2)=0 2x-7=0 or 7x+2=0 2x-7+7=0+7 or 7x+2-2=0-2 2x=7 or 7x=-2 $\frac{2x}{2}$=$\frac{7}{2}$ or $\frac{7x}{7}$=$\frac{-2}{7}$ x=$\frac{7}{2}$ or x=$\frac{-2}{7}$ The solutions are $\frac{7}{2}$ and $\frac{-2}{7}$. check Let x=$\frac{7}{2}$ (2$\frac{7}{2}$ -7)(7$\frac{7}{2}$ +2)= (7-7))(7$\frac{7}{2}$ +2)=0 (0)(7$\frac{7}{2}$ +2)=0 0=0 Let x=$\frac{-2}{7}$ (2x-7)(7x+2)=0 [2($\frac{-2}{7}$)-7][7($\frac{-2}{7}$)+2] [2($\frac{-2}{7}$)-7](-2+2)=0 [2($\frac{-2}{7}$)-7]0=0 0=0
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