SOLUTION: find two consecutive negative integers whose product is 48 more than three times the first integer

SOLUTION: find two consecutive negative integers whose product is 48 more than three times the first integer

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Question 697748: find two consecutive negative integers whose product is 48 more than three times the first integer
Answer by CubeyThePenguin(3113) (Show Source): You can put this solution on YOUR website!
consecutive negative integers: x, (x+1)

x(x+1) = 48 + 3x
x^2 + x = 48 = 3x
x^2 - 2x - 48 = 0
(x - 8)(x + 6) = 0
x = 8, x = -6

x is negative ---> integers: -6, -5
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