請作答^^
更新1:
Quadratic問題來咖喎…希望你地清楚… 我唔知信誰…有人兩個value，有人一個… 跟住要係equal roots，請check數。
我又來問數學= =
回答 (3)
2006-10-16 2:23 am
✔ 最佳答案
(x-1)(x+k)+1=0 has two equal roots, so b^2-4ac = 0(x-1)(x+k)+1 = 0
x^2+(k-1)x+1-k = 0
b^2-4ac = (k-1)^2-4(1)(1-k) = 0
k^2-2k+1-4+4k = 0
k^2+2k-3 = 0
(k+3)(k-1) = 0
k = -3 or k = 1
2006-10-16 2:31 am
Ouesion:(X-1)(X+k)+1=0
(if X is 0)
(0-1)(0+k)+1=0
= -1xk+1=0
= -1k+1=0
= -1k=-1
= k=1
(0-1)(0+1)+1=0
-1x1+1=0
-1+1=0
0=0
Therefor: k is = to 1
(if X is 0)
(0-1)(0+k)+1=0
= -1xk+1=0
= -1k+1=0
= -1k=-1
= k=1
(0-1)(0+1)+1=0
-1x1+1=0
-1+1=0
0=0
Therefor: k is = to 1
2006-10-16 2:20 am
If (x-1)(x+k)+1=0 has two equal roots,find the value of k
請作答^^
x^2+(k-1)x-k+1=0...
因庶equal root...
(k-1)^2-4(1-k)=0
k^2-2k+1-4+4k=0
k^2+2k-3=0
(k+3)(k-1)=0...
k=-3 or1
請作答^^
x^2+(k-1)x-k+1=0...
因庶equal root...
(k-1)^2-4(1-k)=0
k^2-2k+1-4+4k=0
k^2+2k-3=0
(k+3)(k-1)=0...
k=-3 or1
收錄日期: 2021-04-23 13:22:12
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061015000051KK04570