急!!!!!

given that a≠b and α is a common root of the equations x2 +ax+b=0 and x2+bx+a=0. find the value of α and prove that a+b= -1
Let f(x) = x2 + ax + b
As α is a root of f(x) = 0,
So f(α) = 0
α2 + aα + b = 0 .... (1)
Let g(x) = x2 + bx + a
As α is a root of g(x) = 0,
So g(α) = 0
α2 + bα + a = 0 .... (2)
Consider (1) - (2):
α2 + aα + b - (α2 + bα + a) = 0 - 0
(a-b)α + (b-a) = 0
(a-b)α - (a-b) = 0
(a-b)(α-1) = 0
α-1 = 0 【As a≠b, so a-b≠0】
α = 1
Now consider (1) + (2):
α2 + aα + b + (α2 + bα + a) = 0 + 0
2α2 + (a+b)α + (a+b) = 0
2(1)2 + (a+b)(1) + (a+b) = 0
2 + 2(a+b) = 0
a+b = -1.
So a+b = -1.

2007-01-11 12:24:45 補充：
小小補充：這條題目應用到「如果 α 是 f(x) 的根，那麼 f(α) = 0」

given that a≠b and α is a common root of the equations x2 +ax+b=0 and x2+bx+a=0. find the value of α and prove that a+b= -1

Let f(x) = x2 + ax + b

As α is a root of f(x) = 0,

So f(α) = 0

α2 + aα + b = 0 .... (1)

Let g(x) = x2 + bx + a

As α is a root of g(x) = 0,

So g(α) = 0

α2 + bα + a = 0 .... (2)

Consider (1) - (2):

α2 + aα + b - (α2 + bα + a) = 0 - 0

(a-b)α + (b-a) = 0

(a-b)α - (a-b) = 0

(a-b)(α-1) = 0

α-1 = 0 【As a≠b, so a-b≠0】

α = 1

Now consider (1) + (2):

α2 + aα + b + (α2 + bα + a) = 0 + 0

2α2 + (a+b)α + (a+b) = 0

2(1)2 + (a+b)(1) + (a+b) = 0

2 + 2(a+b) = 0

a+b = -1.

So a+b = -1

what!!!!!
find the value ???