SOLUTION: The sum of the squares of three consecutive integers is 110. Find the numbers.

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Question 406783: The sum of the squares of three consecutive integers is 110. Find the numbers.
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!

Hi
three consecutive integers
Let x,(x+1),(x+2) represent the three consecutive integers
Question states***
x^2 +(x+1)^2+(x+2)^2 = 110
Solving for x
3x^2 + 6x + 5 = 110
3x^2 + 6x -105 = 0
3(x^2 + 2x - 35) = 0
(x^2 + 2x - 35) = 0
factoring
(x+7)(x-5) = 0 Note:SUM of the inner product(7x) and the outer product(-5x) = 2x
(x+7)=0 |x = -7 The three consecutive integers are -5,-6,-7
(x-5)=0 |x = 5 The three consecutive integers are 5,6,7

CHECKING our Answer(s)***
25 + 36 + 49 = 110
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