Maths. F4 Quadratic equations (Nature of roots) 15pts.

A. 9x2 - (k+6)x+k - 2 = 0
△=0
[-(k+6)] 2 -4(9)(k-2)=0
(k2+12k+36)-36k+72=0
k2-24k+108=0
k=18 or k=6
B. (k+2)x2 + 6kx + 1 = 0
△=0
(6k) 2-4(k+2)(1)=0
36k2-4k-8=0
9k2-k-2=0
(k-1/18)2=73/324
k-1/18=(√73)/18
k=[1(√73)]/18

C. (x-1) 2 = kx
i.e. x2-2x+1=kx
x2-(2+k)x+1=0
△=0
[-(2+k)] 2-4(1)(1)=0
2k+k2=0
k=0 or k=-2


D. kx2 - (2k+1)x + k = 0
△=0
[- (2k+1)] 2-4(k)(k)=0
4k2-4k+1-4k^2=0
-4k+1=0
k=1/4

E. kx2 - 2(k-1)x + (k+1) = 0
△=0
[- 2(k-1)] 2 -4(k)(k+1)=0
4(k^2-2k+1)-4k2-4k=0
-12k+4=0
k=1/3

F. x^2 + 2(k-4)x + k2 + 6k + 3 = 0
△=0
[2(k-4)] 2 -4(1)(k2+6k+3)=0
4(k2-8k+16)-4k2-24k-12=0
-56k+52=0
k=13/14


2008-08-09 13:01:19 補充：
B有沒有打錯？!

我a同b都算錯答案...唔該幫幫手...唔係好色計

你只要將算式展開,然後找出D,再代D=0,因為the equations has equal roots,就可求出K的值