HKCEE Question (With steps)

2007-12-03 6:09 am

If f(x)=2x^2-3x+4, then f(1)-f(-1)=

If the quadratic equation kx^2+6x+(6-k)=0 has equal roots, then k=

Let x and y be non-zero numbers. If 2x-3y=0, then (x+3y):(x+2y)=

a) 3:2

b)4:3

c)9:7

d) 11:8
A(2,5) and B(6,-3) are two points. If P is a point lying on the straight line x=y such as AP=PB, then the coordinates of P are?

回答 (1)

Chris

2007-12-03 6:23 am

✔ 最佳答案

f(1)-f(-1)=[2(1)^2 -3(1)+4]-[2(-1)^2 -3(-1)+4]
f(1)-f(-1)=2-3+4-2-3-4
f(1)-f(-1)=-6

kx^2+6x+(6-k)=0
b^2-4ac=0
6^2 -4(k)(6-k)=0
36-24k+4k^2=0
k^2-6k+19=0
(k+2)(k-8)=0
k=-2 Or k=8

2x-3y=0
2x=3y
x/y= 3/2
x:y=3:2
(x+3y):(x+2y)=(3+6):(3+4)
(x+3y):(x+2y)=9:7

Let P(k,k)
AP=PB
(k-2)^2 + (k-5)^2= (k-6)^2 +(k+3)^2
k^2-4k+4+k^2-10k+25=k^2-12k+36+k^2+6k+9
-14k+29=-6k+45
8k=-16
k=-2
P(-2,-2)

2007-12-02 22:26:40 補充：
SOR.....第二題睇錯嘢kx^2 6x (6-k)=0 b^2-4ac=06^2 -4(k)(6-k)=036-24k 4k^2=0k^2-6k 9=0(k-3)^2=0k=3

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