(x-2)(x+2) = 9+12x
x = 13
So, the numbers are 11,13,15
Solve algebraically using one variable: Find three consecutive odd integers such that the
product of the first integer and the third integer is equal to nine more than twelve times the middle integer.
3 answers
Consecutive odd have a difference of two.
1st = n 2nd = n + 2 3rd =n + 4
n(n + 4) = 9 + 12(n+2)
n^2 + 4n = 9 + 12n + 24
n^2 -8n -33 = 0
(n -11)(n+3) = 0
n - 11 = 0 n + 3 = 0
n = 11 or n = -3
11, 13, 15 or -3,-1, 0
If you use these choices to do the check; 11, 13 and 15 work.
1st = n 2nd = n + 2 3rd =n + 4
n(n + 4) = 9 + 12(n+2)
n^2 + 4n = 9 + 12n + 24
n^2 -8n -33 = 0
(n -11)(n+3) = 0
n - 11 = 0 n + 3 = 0
n = 11 or n = -3
11, 13, 15 or -3,-1, 0
If you use these choices to do the check; 11, 13 and 15 work.
n (n+2) (n+4)
n(n+4) = 9 +12(n+2)
n^2 + 4 n = 9 + 12 n + 24
n^2 - 8 n - 33 = 0
(n-11)(n+3) = 0
n = 11
n+2 = 13
n+4 = 15
n(n+4) = 9 +12(n+2)
n^2 + 4 n = 9 + 12 n + 24
n^2 - 8 n - 33 = 0
(n-11)(n+3) = 0
n = 11
n+2 = 13
n+4 = 15
1 answer