F.4 Quadratic equation

1. sum of 2 root = 0
=> k^2 - 3 = 0
=> k = sqrt (3) or - sqrt (3)


2. let a and b be the 2 roots of ax^2+bx+c=0
=> 1/a + 1/b = 1
=> (a+b) / ab = 1
=> -b/c = 1
the condition is -b/c = 1

2007-10-14 00:24:14 補充：
when k = -sqrt(3), 4x^2 + (5 + 7sqrt(3)) = 0, 點解 rejected咁 x = sqrt(-(5 + 7sqrt(3)))/2 and -sqrt(-(5 + 7sqrt(3)))/2complex no 都可以 opposite in side

1. Given the equation 4x^2-(k^2-3)x+(2k^2-7k-1)=0, find the value of k such that the two roots are numerically equal but opposite in sign.
m, -m are roots of equation
m + (-m) = (k^2-3)/4 = 0
k = sqrt(3) or k = -sqrt(3)
equation : 4x^2 + (2(3) - 7k - 1) = 0
4x^2 + (5 - 7k) = 0
when k = -sqrt(3), 4x^2 + (5 + 7sqrt(3)) = 0, rejected
so, k = sqrt(3)

2. State the condition that the sum of reciproclas of the roots of the equation ax^2+bx+c=0 is 1.
m,n are roots of equation
1/m + 1/n
= (m + n) / (mn)
= (-b/a) / (c/a)
= -b/c
= 1
so, condition : -b = c

2007-10-13 23:54:59 補充：
慢左, 不過第一題我做得齊 d, 唔該投我一票啦