SOLUTION: If three numbers are consecutive positive integers and five times the square of the largest number is greater than two times the sum of the squares of the other two numbers by 7

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Question 1097613: If three numbers are consecutive positive integers
and five times the square of the largest number is
greater than two times the sum of the squares of
the other two numbers by 75, then the sum of the
smallest and the largest of these numbers is
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I would call the three consecutive positive integers
n-1 , n , and n%2B1 .

Five times the square of the largest number is
5%28n%2B1%29%5E2 .

Two times the sum of the squares
of the other two numbers is
2%28n%5E2%2B%28n-1%29%5E2%29 .

The question states that the first expression above
is greater than the second one by 75.
That can be expressed as
5%28n%2B1%29%5E2=2%28n%5E2%2B%28n-1%29%5E2%29%2B75 .

We can simplify that equation and solve for n .
If we find one or more solutions that is/are positive and integer,
we almost have the answer,
because what we want to find is the sum of the
smallest and the largest of the three consecutive positive integers,
and that would be %28n-1%29%2B%28n%2B1%29=n-1%2Bn%2B1=2n .

Solving %28n%2B1%29%5E2=2%28n%5E2%2B%28n-1%29%5E2%29%2B75 :
5%28n%2B1%29%5E2=2%28n%5E2%2B%28n-1%29%5E2%29%2B75
5%28n%5E2%2B2n%2B1%29=2%28n%5E2%2Bn%5E2-2n%2B1%29%2B75
5n%5E2%2B10n%2B5=2%282n%5E2-2n%2B1%29%2B75
5n%5E2%2B10n%2B5=4n%5E2-4n%2B2%2B75
5n%5E2%2B10n%2B5=4n%5E2-4n%2B77
5n%5E2%2B10n%2B5-4n%5E2%2B4n-77=0
n%5E2%2B14n-72=0
%28n%2B18%29%28n-4%29=0 --> system%28n--18%2C%22or%22%2Cn=4%29
The only possible answer comes from the positive integer n=4 ,
and the answer is 2n=2%2A4=highlight%288%29