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: 15

Answer

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

Work Step by Step

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