Question 1121374: Find three positive consecutive integers such that the product of the first and second is equal to two more than ten times the third integer.
Found 2 solutions by Boreal, FrankM:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! integers are x, x+1, and x+2
x(x+1)=10(x+2)+2
x^2+x=10x+20+2
x^2-9x-22=0
(x-11)(x+2)=0
x=11, only positive result
11,12,13
product of first two is 132, which is 2 more than 10 times the third.
Answer by FrankM(1040)  (Show Source):
You can put this solution on YOUR website! Three consecutive integers.
X, X+1, X+2
(X)(X+1)=10(X+2)+2
X^2+X= 10X+22
X^2-9X-22=0
Use quadratic equation to get te positive result X=11.
Answer is 11,12,13
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