Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 6 - Section 6.6 - Solving Quadratic Equations by Factoring - Exercise Set - Page 473: 52


$y=-5$ or $y=4$

Work Step by Step

$ y(y+9)=4(2y+5)\qquad$...apply the distributive rule: $m(n\pm c)=mn\pm mc$ $ y^{2}+9y=8y+20\qquad$...add $(-8y-20)$ to both sides $y^{2}+y-20=0$ ... Searching for two factors of $ac=-20$ whose sum is $b=1,$ we find$\qquad 5$ and $-4.$ Rewrite the middle term and factor in pairs: $y^{2}+5y-4y-20=0$ $y(y+5)-4(y+5)=0$ $(y+5)(y-4)=0\qquad$...apply the principle of zero products. $y+5=0$ or $y-4=0$ $y=-5$ or $y=4$ Type the equation into a graphing utility and see that the graph intercepts the x-axis at $-5$ and $4$.
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