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 457: 31

Answer

The solutions are -3 and 12.

Work Step by Step

(x+4)(x-9)=4x $x^{2}$-9x+4x-36=4x $x^{2}$-5x-36=4x $x^{2}$-5x-36-4x=4x-4x $x^{2}$-5x-36-4x=0 $x^{2}$-9x-36=0 $x^{2}$+3x-12x-36=0 x(x+3)-12(x+3)=0 (x-12)(x+3)=0 x+3=0 x-12=0 x+3-3=0-3 or x-12+12=0+12 x=-3 or x=12 The solutions are -3 and 12. Check Let x=-3 (x+4)(x-9)=4x (-3+4)(-3-9)=4(-3) 1(-12)=-12 -12=-12 Let x=12 (x+4)(x-9)=4x (12+4)(12-9)=4*12 16*3=48 48=48
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