SOLUTION: Find four consecutive odd integers such that the product of the second and third exceeds the square of the first by 98.

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Question 338997: Find four consecutive odd integers such that the product of the second and third exceeds the square of the first by 98.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
let the integers be x,x+2,x+4,x+6
product of second and third = (x+2)(x+4)
Square of first = x^2
..
x^2+98 = (x+2)(x+4)
x^2+98 = x^2+6x+8
6x+8 = 98
6x =98-8
6x =90
x= 15
.....
15,17 19,21