SOLUTION: given 3 consecutive positive integers, it turns out that the difference between the square of the largest number and the square of the smallest number is 12 less than the square of
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Question 305106: given 3 consecutive positive integers, it turns out that the difference between the square of the largest number and the square of the smallest number is 12 less than the square of the middle number. what is the sum of the 3 numbers? Answer by dabanfield(803) (Show Source):
You can put this solution on YOUR website! given 3 consecutive positive integers, it turns out that the difference between the square of the largest number and the square of the smallest number is 12 less than the square of the middle number. what is the sum of the 3 numbers?
Let x be the samllest of the three consecutive integers. Then the other two are x+1 and x+2. We have then:
(x+2)^ - x^2 = (x+1)^2 - 12
x^2 + 4x + 4 - x^2 = x^2 + 2x + 1 - 12
4x + 4 = x^2 + 2x - 11
x^2 - 2x - 15 = 0
(x-5)*(x+3) = 0
x = 5 or x = -3.
Since the integers are positive x must be 5.
The other two integers are x+1 = 6 and x+2 = 7.
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