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: 33


The solutions are -2 and $\frac{7}{3}$.

Work Step by Step

x(3x-1)=14 3$x^{2}$-x=14 3$x^{2}$-x-14=14-14 3$x^{2}$-x-14=0 3$x^{2}$-7x+6x-14=0 x(3x-7)+2(3x-7)=0 (x+2)(3x-7)=0 x+2=0 3x-7=0 x+2-2=0-2 or 3 x-7+7=0+7 x=-2 or 3 x=7 x=-2 or $\frac{3x}{3}$=$\frac{7}{3}$ x=-2 or x=$\frac{7}{3}$ The solutions are -2 and $\frac{7}{3}$. Check Let x=-2 x(3x-1)=14 -2[3(-2)-1]=14 -2[-6-1]=14 -2[-7]=14 14=14 Let x=$\frac{7}{3}$ x(3x-1)=14 $\frac{7}{3}$(3$\frac{7}{3}$-1)=14 $\frac{7}{3}$(7-1)=14 $\frac{7}{3}$(7-1)=14 $\frac{7}{3}$*6=14 $\frac{42}{3}$=14 14=14
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