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: 56

Answer

The solutions are -2 and 9.

Work Step by Step

(y-5)(y-2)=28 $y^{2}$-2y-5y+10=28 $y^{2}$-7y+10-28=28-28 $y^{2}$-7y+10-28=0 $y^{2}$-7y-18=0 $y^{2}$+2y-9y-18=0 y(y+2)-9(y+2)=0 (y-9)(y+2)=0 y-9=0 or y+2=0 y=9 or y=-2 The solutions are -2 and 9. Check Let x=-2 (y-5)(y-2)=28 (-2-5)(-2-2)=28 -7(-4)=28 28=28 Let x=9 (y-5)(y-2)=28 (9-5)(9-2)=28 4*7=28 28=28
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