急住要架~thxs~有一題數唔識~

a)代x=1, (1)²-5(1)+k=0
1-5+k=0
k=4

b) f(x) = x²-5x+k
= x²-5x+4
= (x-4)(x-1)
x=4 or x=1

c)f(3)=(3)²-5(3)+4=9-15+4= -2

a(-2)=6
a=-3

(a)將1代入(x)的位置, 得 :

f(1) = (1)² - 5(1) + k = 0
1 - 5 + k = 0
- 4 + k = 0
k = 4

(b) x²- 5x + (4)

利用十字相乘法, 得出


= (x-4)(x-1)


(c)a x f(3) = (3) ² - 5 (3) + 4 = 6

a (3) ² - 5(3) + 4 = 6

9a – 15 + 4 = 6

9a – 11 = 6

9a = 17

a = 17/9

a = 1.888888

1(a) 求k的值，
f(1)=(1)^2-5( 1)+k=1-5+k=k-4
f(1)=0
k-4=0
k=4

(b) 解方程f(x)=0
x^2 - 5x+4=0
(x-4)(x-1)=0
x=1, x=4
(c) 求a的值，使a f(3)=6
a*f(3)=6
a*[3^2 -5*3 + 4]=6
a*[9-15+4]=6
a*-2=6
a=-3