SOLUTION: The sum of the squares of three consecutive positive even integers is 116. What are the numbers?

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: The sum of the squares of three consecutive positive even integers is 116. What are the numbers?      Log On


   

Question 414094: The sum of the squares of three consecutive positive even integers is 116.
What are the numbers?
Found 2 solutions by Alan3354, ewatrrr:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
The sum of the squares of three consecutive positive even integers is 116.
What are the numbers?
------------------
116/3 = apx 39
sqrt(39) = apx 6, the middle number
--> 4, 6 & 8

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

Let x,(x+2),(x+4)represent the three consecutive positive even integers 

Question states***

  x^2 + (x+2)^2 + (x+4)^2 = 116

Solving for x

  3x^2 + 12x + 20 = 116

  3x^2 + 12x - 96 = 0

   x^2 + 4x - 32 = 0

factoring

 (x + 8)(x-4)=0  Note:SUM of the inner product(8x) and the outer product(-4x) = 4x

 (x + 8)=0  x = -8 is an Extraneous solution (not 'positive')

 (x-4)=0    x = 4  the three consecutive positive even integers are 4,6,8



CHECKING our Answer***

 16 + 36 + 64 = 116