假設二次方程 x^2 - 5x + k (1-x)=0 的 一個根為 2
(a) 搵k的值
(b) 搵方程的另一個根
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2007-09-27 8:11 pm
回答 (2)
2007-09-27 8:18 pm
✔ 最佳答案
2^2-5(2)+k(1-2)=04-10-k=0
4-10=k
-6=k
x^2-5x-6(1-x)=0
x^2-5x-6+6x=0
x^2+x-6=0
(x+3)(x-2)=0
x= -3 / x = 2
so x = -3
2007-09-27 8:40 pm
a)
Put 2 into the equation
2^2 -5(2) + k(1-2) = 0
k = -6
b)
The equation is x^2 +x -6 = 0
Dividing (x^2 +x -6) by (x-2) gives (x+3)
The equation becomes (x-2)(x+3)=0
The other root is -3
Put 2 into the equation
2^2 -5(2) + k(1-2) = 0
k = -6
b)
The equation is x^2 +x -6 = 0
Dividing (x^2 +x -6) by (x-2) gives (x+3)
The equation becomes (x-2)(x+3)=0
The other root is -3
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