f.4 a math超急(1)!!

let f(x)=2x2+2(k-4)x+k,wher k is real .
(A)find the discriminant of the equation f(x)=0
B2-4AC
= [-2(k-4)]2-4(2)(k)
= 4k2-32k+64-8k
= 4k2-40k+64
= 4(k2-10k+16)
= 4(k-2)(k-8)

(b)if the graph y=f(x)lies above the x-axis for all values of x,fing the range of possible values of k
B2-4AC > 0
4(k-2)(k-8) > 0
so that
k > 8 or k < 2


2006-11-02 22:32:58 補充：
sooryB2-4AC 2

2006-11-02 22:35:16 補充：
sorry something wrongB^2-4AC＜2such thatk ＜ 8 and k ＞ 2

A.

f(x)=2x^2+2(k-4)x+k = 0

discriminant = [2(k-4)]^2 - 4*(2)*(k)
=4k^2 - 16k + 32 - 8k
=4k^2 - 24k + 32
=4(k^2 - 6k + 8)

B.

if all values of x, the graph y=f(x) lies above x-axis, y=f(x) is greater than zero.
So, there is no solution for y=f(x)=0.
So, the discriminant must be less than zero.

4(k^2 - 6k + 8) less than zero
(k^2 - 6k + 8) less than zero
(k-2)(k-4) less than zero
k is within 2 and 4.