SOLUTION: Fidn three consecutive odd integers such that the sum of all three is 42 less than the product of the larger two.

SOLUTION: Fidn three consecutive odd integers such that the sum of all three is 42 less than the product of the larger two.

Algebra.Com

Question 864318: Fidn three consecutive odd integers such that the sum of all three is 42 less than the product of the larger two.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
let the integers be x x+2 & x+4
x+x+2+x+4 = (x+2)(x+4)-42
3x+6=x^2+6x+8-42
3x+6 =x^2+6x-40
x^2+3x-40=0
x^2+8x-5x-40=0
x(x+8)-5(x+8)=0
(x+8)(x-5)=0
x=-8 OR x=5
but odd integer
so, x=5
5,7,9

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