F.4 A-maths

1) 5x^2+10x+k
= 5(x^2 + 2x) + k
= 5(x + 1)^2 + k - 5
because 5(x + 1)^2 is greater than 0
so,, the least value of k = 5..

4) 同樣方法...completing square..
(m^2.x^2)-mx+1
= m^2 ( x^2 - x/m ) + 1
= m^2 [ x - 1/(2m)] ^2 + 1 - [m^2 / (4m^2)]
= m^2 [ x - 1/(2m)] ^2 + 1 - 1/4
= m^2 [ x - 1/(2m)] ^2 + 3/4
so,, m can be any values...because m^2 is always positive.

2) 都係completing square..自己試下..
(2a) 答案應該係18/16
(2b) 答案應該係 20 and it must be greater than 0.

3) 唔想計了..同樣方法@@

1. k<5
2.a) k<9/8
b) k<4/5
3.k>64
4.m>0
如果错了不要怪我哦!

**words inside ( ) are explantions only. dont copy them**

1. Discriminant > 0
10^2 - 4(5)(k) > 0
100>20k
k<5

2a. Discriminant > 0
3^2 - 4(2)(k) > 0
9-8k > 0
k < 9/8

b. Discriminant > 0, k > 0
(-4)^2 - 4(k)(5) > 0
16 - 20k > 0
k < 4/5

3. (24-x)(x-8)-k
= 24x - 192 - x^2 + 8x - k
= -x^2 + 32x - 192 - k
Discriminant > 0
32^2 - 4(-1)(-192-k) > 0
1024 - 768 - 4k > 0
256 > 4k
k < 64

4.Let (m^2.x^2)-mx+1 be f(x)
f(x) = (mx)^2 - mx +1
(By using the completing the square method)
= (mx - 1/2)^2 - 1/4 + 1
= (mx - 1/2)^2 + 3/4
f(x) > 3/4
f(x) > 0
(m^2.x^2)-mx+1 > 0

1. 6
2a. 2
2b. 1
3. 65
4. m^2(x-1/2m)^2+3/4