f.4 Maths (Quadratic equation)

1)
(4-x)^2-5(4-x)-14=0
[(4-x)-7][(4-x)+2]=0
4-x=7 or 4-x=-2
x=-3 or x=6
2)
4(x-3)^2-100=0
4(x-3)^2=100
(x-3)^2=25
x-3=5 ot x-3=-5
x=8 or x=-2
3)
(3x+1)(1-2x)=-4
3x-6x^2+1-2x=-4
6x^2-x-5=0
(6x+5)(x-1)=0
x=-5/6 or x=1


2009-09-10 19:23:25 補充：
6x^2-x-5=0

(6x^2+5x)(-6x-5) <--你呢度錯左
你唔應該咁寫...你只是寫成(6x²+5x)+(-6x-5)
中間的關係不是product形式

你好，希望能夠幫到你
1.(4-x)^2-5(4-x)-14=0
(x^2 - 8x + 16) - (20 + 5x) - 14 = 0
x^2 - 3x - 18 = 0
(x - 6) (x + 3) = 0
x = 6 或 x = -3

2. 4(x-3)^2-100=0
4(x^2 - 6x +9 ) -100=0
4x^2 - 24x + 36 - 100= 0
(x - 8)(x + 2) = 0
x= 8 或 x=-2

3. (3x+1)(1-2x)=-4
(3x+1)(-2x+1)+4 = 0
-6x^2 + x +5 = 0
(6x+5) (x-1)=0
x=-5/6 或 x= 1