Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 9 - Quadratic Functions and Equations - 9-4 Factoring to Solve Quadratic Equations - Practice and Problem-Solving Exercises - Page 558: 13


The solutions are $1\frac{3}{4}$ and $-2\frac{2}{3}$.

Work Step by Step

If a product is 0, one of the factors must be 0. Use the addition property of equality and the multiplication property of equality to solve for a. $4a-7=0\ or\ 3a+8=0$ $4a-7+7=0+7\ or\ 3a+8-8=0-8$ $4a=7\ or\ 3a=-8$ $4a\div4=7\div4\ or\ 3a\div3=-8\div3$ $a=\frac{7}{4}\ or\ a=-\frac{8}{3}$ $a=1\frac{3}{4}\ or\ a=-2\frac{2}{3}$
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