a maths

α + β = k + 2
αβ = k

(α + 1)(β + 2) = 4
αβ + 2α + β + 2 = 4
k + α + k + 2 + 2 = 4
α + 2k = 0
α = -2k

-2k + β = k + 2
β = 3k + 2

αβ = k
(-2k)(3k + 2) = k
-6k^2 - 5k = 0
k(6k + 5) = 0

k = 5/6, k = 0

2007-01-01 04:36:35 補充：
漏左負號k = -5/6, k = 0

α, β are the roots of the quadratic equation x2-(k+2)x+k=0.
from a x2 +b x + c = 0, a=1, b=-(k+2) , c=k
=> sum of root α+ β = -b/a = - [-(k+2)] = k+2
product of root αβ = c/a = k
if (α+1)(β+2) = 4,show that α= -2k.
given (α+1)(β+2) = 4
αβ + 2α + β+ 2 = 4
so , αβ + (α + β) + α = 2
k + k+2 +α = 2
2k + α = 0
therefore α= -2k (proved)

Hence, α is one of the root of the equation,
We put x=α= -2k into the equation,
We get (-2k)2-(k+2)(-2k)+k=0
4k2 + 2k2 + 4k + k =0
6k2 + 5k = 0
k (6k + 5) = 0
k = 0 or k = -5/6