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 - Page 458: 35


The solutions are $\frac{8}{3}$ and -9.

Work Step by Step

3$x^{2}$+19x-72=0 3$x^{2}$+27x-8x-72=0 3x(x+9)-8(x+9)=0 (3x-8)(x+9)=0 (3x-8)=0 or (x+9)=0 3x-8+8=0 +8 or x+9-9=0-9 3x=8 or x=-9 $\frac{3x}{3}$ =$\frac{8}{3}$ or x=-9 x= $\frac{8}{3}$ or x=-9 The solutions are $\frac{8}{3}$ and -9. Check Let x=$\frac{8}{3}$ 3$x^{2}$+19x-72=0 3$(8/3)^{2}$+19$\frac{8}{3}$-72=0 3$\frac{64}{9}$+$\frac{152}{3}$-72=0 $\frac{64}{3}$+$\frac{152}{3}$-72=0 $\frac{216}{3}$ -72=0 72-72=0 0=0 Let x=-9 3$x^{2}$+19x+19(-9)-72=0 3$(-9)^{2}$ +19(-9)-72=0 3*81-171-72=0 243-171-72=0 0=0
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