SOLUTION: Three consecutive even integers are such that the square of the third is 100 more than the square of the second. Find the three integers.

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Question 191906: Three consecutive even integers are such that the square of the third is 100 more than the square of the second. Find the three integers.
Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
let x, (x+2) & (x+4) be the three consecutive even integers.
(x+4)^2=(x+2)^2+100
x^2+8x+16=x^2+4x+4+100
x^2-x^2+8x-4x=100+4-16
4x=88
x=88/4
x=22 ans. for the first integer.
22+2=24 for the middle integer.
22+4=26 for the third integer.
Proof:
26^2=24^2+100
676=576+100
676=676