Algebra: A Combined Approach (4th Edition)

Published by Pearson

Chapter 6 - Section 6.6 - Solving Quadratic Equations by Factoring - Exercise Set - Page 458: 55

Answer

The solutions are -4 and 3.

Work Step by Step

(y-2)(y+3)=6 $y^{2}$+3y-2y-6=0 $y^{2}$+y-6+6=6+6 $y^{2}$+y=12 $y^{2}$+y-12=0 $y^{2}$+4y-3y-12=0 y(y+4)-3(y+4)=0 (y-3)(y+4)=0 y-3=0 or y+4=0 y=3 or y=-4 The solutions are -4 and 3. Check Let x=-4 (y-2)(y+3)=6 (-4-2)(-4+3)=6 -6(-1)=6 6=6 Let x=3 (y-2)(y+3)=6 (3-2)(3+3)=6 1*6=6 6=6

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