SOLUTION: find three consecutive even integers such that the product of the first and third is 18 greater than the product of -1 and the third.

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: find three consecutive even integers such that the product of the first and third is 18 greater than the product of -1 and the third.      Log On


   

Question 1152224: find three consecutive even integers such that the product of the first and third is 18 greater than the product of -1 and the third.
Found 2 solutions by josgarithmetic, MathTherapy:
Answer by josgarithmetic(39628) About Me  (Show Source):
You can put this solution on YOUR website!
Even integers n, n+2, n+4

mistake in first step
n%28n%2B4%29=18%2B%28-1%29%28cross%28n%29%28n%2B4%29%29

n%28n%2B4%29=18-%28n%2B4%29
n%5E2%2B4n=18-n-4
n%5E2%2B5n=14
n%5E2%2B5n-14=0
%28n%2B7%29%28n-2%29=0

n=-7 will not be for even integers.

n=2 will give even integers.
---------------------------
---------------------------
the integers: 2, 4, 6
---------------------------
---------------------------

Answer by MathTherapy(10556) About Me  (Show Source):
You can put this solution on YOUR website!
find three consecutive even integers such that the product of the first and third is 18 greater than the product of -1 and the third.
Let 1st integer be F

Then the 2nd and 3rd are: F + 2, and F + 4, respectively

Then we get the following equation: F(F + 4) = - 1(F + 4) + 18



(F - 2)(F + 7) = 0

F, or 1st integer = 2             OR              F = - 7 (ignore)

You should be able to list the other 2 now!

Note that the other person is WRONG, as usual!