F.4 maths *c

10.
a)
Since the given quadratic equation has real roots, the discriminant is great than or equal to 0.
(2k)^2 - 4(k+1)(k-2) > = 0
4k^2 - 4(k^2-k-2) > = 0
4k^2 - 4k^2 + 4k + 8 >= 0
4k + 8 >= 0
4k >= -8
k >= -2

b)
(i)
For k = -2,f(x)=0 becomes -x^2 -4x -4 =0.
x^2 +4x +4 =0.
(x+2)^2 = 0
x = -2 (repeated)

(ii)
Let the roots of f(x)=0 be α and β.
α + β: -2k/(k+1)
α β: (k-2)/(k+1)

Let the roots of the required equation be (1/α) and (1/β)
(1/α)+(1/β): (α + β)/(α β) = -2k/(k-2)
(1/α)(1/β): 1/(α β) = (k+1)/(k-2)

The required equation is
x^2 + [2k/(k-2)] x + [(k+1)/(k-2)] = 0
(k-2)x^2 + 2kx + (k+1) = 0

If there is a mistake, please infrom me!

2007-10-23 00:15:07 補充：
If there is a mistake, please inform me!

f(x)=(k+1)‧x^2 +2kx+(k-2) = 0 has real roots.
(a) Discriminant = (2k)^2 - 4(k+1)(k-2) >= 0
4k^2 - (4k^2 - k -2) >0
k+2 >= 0
k>=-2

(b)(i) For k>=-2, the minimum value of k is -2.
f(x) = -x^2-4x-4 = 0
x^2 + 4x+4 = 0
(x+2)^2 = 0
x= -2
(ii) The roots required are -1/2 and -1/2
Hence the equation is (x+1/2)^2 = 0
x^2 + x + 1/4 = 0
i.e. 4x^2 + 4x + 1 = 0