SOLUTION: find 3 positive consecutive intergers such that the product of the first and seconds is 2 more than 9 times the third.

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: find 3 positive consecutive intergers such that the product of the first and seconds is 2 more than 9 times the third.       Log On


   

Question 440820: find 3 positive consecutive intergers such that the product of the first and seconds is 2 more than 9 times the third.
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
Let n be the lowest integer
Then the 3 integers are: n, n+1, n+2
Given: 1st times 2nd is 2 more than 9 times the 3rd.
In equation form this is:
n(n+1) = 9(n+2) + 2
Solve for n:
n^2 + n = 9n + 18 + 2
n^2 - 8n - 20 = 0
Use the quadratic formula:
n = (8 +- sqrt(64 + 80))/2
This gives n = 10, -2
Since the integers are positive, n = 10
So the 3 integers are:
10, 11, 12
Check:
10*11 = 110
9*12 + 2 = 110