F.4 A.Maths Quadratic Equation (10 pts)

Let x2 - 4x+k=0..........(1 )
2x2 - 3x+k=0.........(2)

Subsititute x = a into both equations,
we have:
a2 - 4a = 2a2 - 3a
a2 + a = 0
a = 0 or a = -1

put x = 0 into (1) and (2),
k = 0

put x = -1 into (1),
1 + 4 + k = 0
k = -5

2007-11-11 14:16:49 補充：
Hence,the values of k is 0 or -5

If α is a common roots of x^2-4x+k=0 and 2x^2-3x+k=0 ,
then α^2 - 4α + k = 0 ...(1)
and 2α^2 - 3α + k = 0 ...(2)
(1) x2 - (2): 2α^2 - 8α + 2k - 2α^2 +3α - k = 0
-5α + k = 0
k = 5α ...(3)
Substitute (3) into (1)
α^2 - 4α + 5α = 0
α^2 + α = 0
α(α+1) = 0
α = 0 or -1
When α = 0, k = 5(0) = 0
When α = -1, k = 5(-1) = -5
Hence k = 0 or -5