F.4 二次方程

(x-1)(x-k)+1=0

x^2 - x - kx + k + 1 = 0

x^2 - (k+1)x + (k+1) = 0

兩根和 = (k+1)

由於兩根相等,重根 = (k+1)/2

代入原方程 :

[(k+1)/2]^2 - (k+1) (k+1)/2 + (k+1) = 0

[(k+1)^2 ]/ 4 - [(k+1)^2] /2 + (k+1) = 0

k + 1 = [(k+1)^2 ] / 4

4k + 4 = k^2 + 2k + 1

k^2 - 2k - 3 = 0

(k + 1)(k - 3) = 0

k = - 1 or k = 3





2009-12-25 12:48:46 補充：
我蠢了!

用 b^2 - 4ac = 0 超快

△ = (k+1)^2 - 4(k+1) = 0

(k+1)[(k+1) - 4]= 0

k+1 = 0 or (k+1) - 4 = 0

k = -1 or k = 3

你好醒啊!!

2009-12-26 13:59:19 補充：
z = kx^2 / y

81 = 9k / 2

k = 18

i.e. z = 18x^2 /y

若x增加4%,設y 百分變化 = p%使z增加35.2% :

1.352z = 18(1.04x)^2 /y(p%)

1.352z = (18x^2 /y) * 1.04^2 / (p%)

1.352 (p%) = 1.04^2

p% = 80%

y 減少 20%.

2009-12-26 14:02:37 補充：
2.g(x)=6x-x^2
f(x)=3x+4/x
解方程g (x) = f(x)

6x - x^2 = 3x + 4/x

6x^2 - x^3 = 3x^2 + 4

x^3 - 3x^2 + 4 = 0

(x+1)(x^2 - 4x + 4) = 0

(x+1)(x-2)^2 = 0

x = - 1 or 2(double roots)

2009-12-26 14:22:42 補充：
餘式定理,
(-1)^3 - 3(-1)^2 + 4 = 0

so (x - -1) = x + 1 is a factor of the eq..

用邊條式可以計到x^3

2009-12-26 14:35:07 補充：
除左餘式定理, 仲有冇其他式??