SOLUTION: find three consecutive positive odd integers such that the square of the first is 9 more than 6 times the sum if the second and third.

Question 986806: find three consecutive positive odd integers such that the square of the first is 9 more than 6 times the sum if the second and third.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
find three consecutive positive odd integers
n, n+2, n+4
such that the square of the first is 9 more than 6 times the sum if the second and third.
n^2 = 6((n+2) + (n+4)) + 9
n^2 = 6(2n + 6) + 9
n^2 = 12n + 36 + 9
n^2 = 12n + 45
arrange as a quadratic equation
n^2 - 12n - 45 = 0
Factors to
(n-15)(n+3) = 0
positive solution
n = 15
:
The 3 integers: 15, 17, 19
:
:
See if that works
15^2 = 6(17+19) + 9
225 = 6(36) + 9
225 = 216 + 9
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