maths 高手請入!急!急!

x^2+kx-k+2=0 [α,β]
a)
α+β = -k/1 = -k
αβ = (-k+2)/1 = -k+2
b)i)
(α-1)(β-1) = αβ - α - β + 1 = αβ - (α + β) + 1 = -k+2+k+1 = 3
b)ii)
Since α and β are positive integers,
3 = 3 x 1 = (4-1)(2-1)
As α<β, α=2, β=4
c)
α+β = -k
k = -6

a) Since if a and b are the roots of x^2+kx-k+2=0, then
(x-a)(x-b)=0, then
x^2-(a+b)x+ab=0
then, a+b=-k, ab=2-k
b) (a-1)(b-1)=ab-(a+b)+1
=2-K-(-k)+1=3
ii)since a & b are integer
so, a-1=3 and b-1=1 , or a-1=1 and b-1=3, then
a=4 and b=2, or a=2 and b=4.
c)k=-6
QED. (Quite Easy Done) 哈哈.
sorry, 我唔識打ALPHA, 同BEAT所以用A & B代替

a)
α+β=-k...(1)
αβ=2-k...(2)

bi)
(α-1)(β-1)=αβ-(α+β)+1=2-k-(-k)+1=3

ii)
From the result of bi, since α and β are positive interegers with α<β and the only combination of product of 3 is 1*3, hence we can easily find that α=2 and β=4.

c)
From (1),
k=-(α+β)=-(2+4)=-6