Maths Problem-- Quadratic Equation 緊急

α is a common root of both equations implies:
α^2 + a α + b = 0 …. (1)
α^2 + b α + a = 0 …. (2)
(1) – (2) gives:
(a – b) α + b – a = 0
(a – b) α – (a – b) = 0
(a – b) (α – 1) = 0 …. (3)
Since a is not equal to b, then from (3), we know that α – 1 = 0, so α = 1.
(1) + (2) gives:
2 α^2 + a α + b α + a + b = 0
Since α =1, we have
2 + a + b + a + b = 0
=> 2 (a + b) = -2
=> (a + b) = -1

Hope it is not too late
圖片參考：http://hk.yimg.com/i/icon/16/2.gif

Let the number be xy
Then xy - x - y = (10x + y) - x - y = 9x
So it is always divisible by 9