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Question 759799: Three consecutive positive integers are such that the square of their sum exceeds the sum of their squares by 94. Find these three integers.
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! Let the 3 integers be n, n+1 and n+2
The square of their sum is (n+n+1+n+2)^2 and this is larger than the sum of their squares n^2 + (n+1)^2 + (n+2)^2 by 94
In equation form this is
(n+n+1+n+2)^2 = n^2 + (n+1)^2 + (n+2)^2 + 94
If you perform the multiplication, collect terms and simplify you should get:
n^2 + 2n - 15 = 0
Factor:
(n-3)(n+5) We are told the integers are positive, so the solution is n=3
Ans: 3,4,5
Check:
12^2 = 9 + 16 + 25 + 94 -> 144 = 50 + 94 = 144
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