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 - Lesson Check: 3

Answer

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

Work Step by Step

Factor the equation, and then use the zero product property. (If the product is 0, one of the factors is zero.) $3y^2-17y+24=0$ $(3y-?)(y-?)=0$ $\longrightarrow$ The 2nd term is negative and the 3rd is positive, so the second terms in the factored equation are both negative. $(3y-8)(y-3)=0$ $3y-8=0$ OR $y-3=0$ $\longrightarrow$ solve for t using addition property of equality $3y-8+8=0+8$ OR $y-3+3=0+3$ $3y=8$ OR $y=3$ $3y\div3=8\div3$ OR $y=3$$\longrightarrow$ use the multiplication property of equality $y=\frac{8}{3}$ OR $y=3$ $y=2\frac{2}{3}$ OR $y=3$
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