Question 266103: find three consecutive positive integers such that the product of the first and the third is 29 more than the second. Answer by roseo(33) (Show Source):
You can put this solution on YOUR website! Let x= 1st consecutive integer
x+1 = 2nd consecutive integer
x+2 = 3rd consecutive integer
The product of the first and the third would be x(x+2) and 29 more that the second would be 29+x+1
Then your equation would be x(x+2) = 29+x+1
x^2 +2x=30+x
Put the terms on one side and factor the quadratic equation x^2+x-30=0
(x+6)(x-5)=0
x+6=0 x-5=0
x=-6 and x=5
Because your answers must be positive they would be 5,6, and 7.
(