一元二次方程(4)

6)方程x(x-k)=x-16有一個二重實根
a)求k可能的值
x(x - k) = x - 16
x^2 - kx = x - 16
x^2 - (k+1)x + 16 = 0
delta = (k+1)^2 - (4)(16) = 0
k^2 + 2k - 63 = 0
(k - 7)(k + 9) = 0
k = 7 or k = -9

b)若k<0,解方程
k = -9
x^2 + 8x + 16 = 0
(x + 4)^2 = 0
x = -4

7)考慮二次方程kx^2-kx-5(x-1)=0,其中k不等於0
a)求方程的判別式
kx^2 - kx - 5(x - 1) = 0
kx^2 - (k+5)x + 5 = 0
判別式
= (k+5)^2 - (4)(k)(5)
= k^2 - 10k + 25

b)若方程有兩個等實根,求k
k^2 - 10k + 25 = 0
(k - 5)^2 = 0
k = 5

c)寫出方程的根的性質
5x^2 - 10x + 5 = 0
x^2 - 2x + 1 = 0
(x - 1)^2 = 0
x = 1

8)若y=(k+1)x^2-6x+1與x軸接觸於P
a)求k
(k+1)x^2 - 6x + 1 = 0
delta = 36 - 4(k+1) = 0
k = 8

b)求P的坐標
9x^2 - 6x + 1 = 0
(3x - 1)^2 = 0
x = 1/3
P = (1/3,0)