Question 373769: Find three consecutive integers such that the square of the sum of the smaller two is 105 more than the square of the largest. I would like to know how to solve because I already have the answer which is 6,7,8.
Found 2 solutions by amoresroy, ewatrrr:
Answer by amoresroy(361)   (Show Source):
You can put this solution on YOUR website! Find three consecutive integers such that the square of the sum of the smaller two is 105 more than the square of the largest. I would like to know how to solve because I already have the answer which is 6,7,8.
Let x = the smallest integer
x+1 = the next integer
x+2 = the larget integer
(x+x+1)^2 = (x+2)^2 + 105
(2x+1)^2 = (x+2)^2 + 105
4x^2+4x+1 = x^2+4x+4+105
4x cancels out
4x^2-x^2 = 105+4-1
3x^2 = 108
x^2 = 36
x = 6
x+1 = 7
x+2 = 8
Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website!
Hi,
Let x represent the smallest of the three consecutive integers
then (x+1) and (x+2) would be the next two consecutive integers
Write as we Read
[x + (x+1)]^2 = (x+2)^2 +105
solving for x
[2x + 1)]^2 = (x+2)^2 + 105
4x^2 + 4x + 1 = x^2 + 4x + 4 +105
3x^2 - 108 = 0
x^2 = 108/3
x^2 = 36
x = +- 6
three consecutive integers are 6,7,8 Or -6,-5,-4
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