中四A. maths問題

x^2 +(k-2)x-(k-1)=0 ; , 〡a〡=〡B〡

Find k

題目given 〡a〡=〡B〡
首先拆左個absolute value 就簡單好多
即係a=B or a = -B

咁表示x^2 +(k-2)x-(k-1)=0 哩條equation 既roots 有兩個可能:
情況1, 有double roots, 即係兩個roots 都等於a
這時, 用sum of roots 同product of roots, 列出兩條式就計到k 既值

情況2, 有two real roots, 即係兩個roots 都等 a 同埋 -a
這時, 用sum of roots 同product of roots, 列出兩條式就計到 k 既值

計法大約係咁, 不過答案係要列晒兩個情況既k 出黎喎.

2007-10-21 10:36:13 補充：
樓上個想法爭d, 佢冇諗到a =-B 果個condition.

2007-10-21 15:51:25 補充：
x^2 +(k-2)x-(k-1)=0case 1:when there is a double roots,sum of roots :a+a =-(k-2)2a =-(k-2)a =-(k-2)/2 ...(1)product of roots :a*a =-(k-1)a^2 =-(k-1) ...(2)sub(1) into (2)[-(k-2)/2]^2 =-(k-1)(k-2)^2/4 =-(k-1)(k-2)^2 =-4(k-1)k^2-4k+4 =-4k+4k^2 =0k=0

2007-10-21 15:52:29 補充：
case 2:when the roots are a and -asum of roots :a (-a) =-(k-2)0 =-(k-2)(k-2)=0k=2其實呢個答案仲直接, 用sum of roots已經攪掂這題冇要求搵埋個root a出黎, 但你自己可以代番入條式度搵下, 當做下exercise.

因為〡a〡=〡B〡
the root is equal
only one real root
b^2-4ac = 0
(k-2)^2 - 4 (1) (-(k-1)) =0
k^2-4k+4 +4k-4 =0
k^2 =0
k=0