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.

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) (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
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