Question

como resolver (3x-1)2-3(2x+3)2+42=2x(-x-5)-(x-1)2

Answer 1

(3x-1)² - 3(2x+3)² + 42 = 2x(-x-5) - (x-1)²

9x² - 6x +1 - 3(4x² +12x + 9) + 42 = -2x² -10x - (x² -2x +1)

9x² - 6x + 1 - 12x² - 36x - 27 + 42 = -2x² -10x - x² +2x -1

9x² - 12x² + 2x² + x² - 6x - 36x + 10x - 2x  + 1 - 27 + 42 + 1 = 0 

-34x + 17 = 0 

-34x = -17 *(-1)

34x = 17 

x = 17/34

x = 1/2

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14/10/2016
Sepauto 
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