f.4 maths 急!!!!!

x^2+2kx+(6+k)=0

(a)Δ≧0

4k^2-4(6+k)≧0

k≧3, k≦-2

(b)equationx^2+2kx+(6+k)=0 has positive roots when k≦-2

positive roots
即係equation既根必須為正數

若方程式has real roots，則
判別式(delta)大於等於零 (greater than or equal to zero)
b^2 – 4ac ≧ 0
(2k)^2 – 4(1)(6+k) ≧ 0
4k^2 – 24 – 4k ≧ 0
k^2 – k – 6 ≧ 0
(k+2)(k-3) ≧ 0
k ≦ -2 or k ≧ 3

(b)positive roots ( I think you mean both roots are positive, not only one root is positive. And I think you mean positive real roots)
sum of roots and product of roots should be positive
-2k > 0 and 6+k > 0 (A maths teach you sum and product of roots)
k < 0 and k > -6
-6 < k ＜ 0
Combining with (a) result (because roots are real)
-6 < k ＜ 0 and (k ≦ -2 or k ≧ 3)
so -6< k ≦ -2



positive roots (I think you are asking positive real roots)，即正數的實根, x的值為正數

x^2+2kx+(6+k)=0

(a) To have real roots, determinant must be greater or equal to 0
Hence, (2k)^2-4(6+k) is greater or equals to 0
k^2-k-6 is greater or equals to 0
k is smaller or equals to -2 or k is greater or equals to 3

(b) To have possible roots, Sum of roots and products of roots are both greater than 0

Sum of roots = -2k is greater than 0
so, k is smaller than 0
products of roots = 6+k is greater than 0
k is greater than -6

So, -6 細過 x 細過 0
再+埋a個結果
因為positive root (正數既根) 一定要係real root
所以 -6 細過 x 細過 -2就係答案

2006-10-26 00:22:39 補充：
大家有無計漏左少少野呢就係兩個係負數相乘都係織正數要2個都係正數除左要計佢地相乘係正數之外都要睇埋佢地相加係唔係正數如果都係就會確定到答案

find the range of real values of k if the quadratic equation x2+2kx+(6+k)=0 has:
(a)real roots;
若方程式有實根，則
判別式大於等於零
B^2 – 4AC ≧ 0
(2k)^2 – 4(1)(6+k) ≧ 0
4k^2 – 24 – 4k ≧ 0
k^2 – k – 6 ≧ 0
(k+2)(k-3) ≧ 0
k ≦ -2 或 k ≧ 3

(b)positive roots;
有正值的根
則
若方程式為 X^2 + BX + C 的形式，它有正值的根，則判別式大於零(有實根)及B為負值，或(B為正值時，C為負值)

B為負值
2k ＜ 0，則k ＜0
或
B為正值時，C為負值
2k ＞ 0 及 6 + k ＜ 0
若 k ＞ 0 則 6 + k 必為正數，所以這條件不成立。
因此要有正及實的根，
所以
k ＜ 0 及 (k ≦ -2 或 k ≧ 3)
所以結果
k ≦ -2

仲有以中文講解positive roots点解
positive roots，即它的根為正值(即正數的實根)。

(a) has real root 即係代表 B2-4AC>=0
(b) positive roots 即係代表 個root係正數
咁即係用quadratic formula 將寫番個roots 跟住就 寫成>0
明嗎?

歡迎 傳呼我問數: 79127456 (高SIR) 但係講明你係學生想問數喎~