Question 474925: find three consecutive integers such that the square of the sum of the smaller two is equal to twice the largest.
(three consecutive integers are x, x+2, and x+4)
(the square of the smaller two are x^2 and (x+2)^2)
Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! three consecutive even integers are x, x+2, and x+4
the square of the sum of smaller two
sum of smaller two = x+x+2 = 2x+2
square = (2x+2)^2
(2x+2)^2= 2(x+4)
4x^2+8x+4=2x+8
4x^2+6x-4=0
4x^2+8x-2x-4=0
4x(x+2)-2(x+2)=0
(4x-2)(x+2)=0
x=1/2 OR -2
eliminate x=1/2 since it is not an integer
(-2,0,2) are the integers
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