Question 318736: Find three consecutive integers such that the square of the sum of the smaller two is 297 more than the square of the largest.
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! let the integers be x, x+1, x+2
..
(x+x+1)^2 = (x+2)^2 +297
(2x+1)^2=x^2+4x+4+297
4x^2+4x+1= x^2+4x+301
3x^2-300=0
3x^2=300
x^2=300/3
x^2=100
x=10
the integers are 10,11,12
|
|
|