(a+1)(b+2)=4 ,show that a= -2k

a+b = k+2
ab = k
(a+1)(b+2)=4
ab+ (a+b) + a + 2 = 4
k + k+2 + a + 2 = 4
2k+ a = 0
a = -2k
Hence x = -2k is one of the roots of the equation.
i.e. (-2k)^2 - (k+2)(-2k) + k = 0
4k^2 + 2k^2 + 4k + k = 0
k(6k+5) = 0
k = 0 or -5/6

given that a+b = k+2 and ab=k

(a+1)(b+2) = 4

Step 1:

a+b = k+2
b = k+2-a

Step 2:


ab+2a+b+2 = 4

ab+2a+b = 4-2

As ab=k and b = k+2-a

k + 2a+ (k+2-a) = 2

2k+a+2 = 2

2k+a+(2-2) = 0

2k+a = 0

a = -2k

Hence a = -2k

2007-11-03 20:35:55 補充：
Step 3 find the two values of k:(a,b) = (x,y) == a=x= -2kx^2 -(k 2)x k=0(-2k)^2 -(k 2)(-2k) k=04k^2 -((-2k)^2-4k) k =04k^2 2k^2 4k k = 0k(6k 5) = 0k = 0 or (6k 5) = 0 == k=-5/6Hence the two values of k are k = 0 or -5/6