α, β are the real roots of the equation x^2+ (k-2)x-(k-1)=0.
If | α | = | β |, find k
quadratic equation
2006-10-17 4:02 am
回答 (4)
2006-10-17 4:05 am
✔ 最佳答案
α 與 β 是real roots of the eqationx2 + ( k-2 ) x - (k -1) = 0
如果|α| = |β| , --------( absolute α= absolute β )
請找 k的數值。
因
若α = -β，則
α + β = 0，
k – 2 = 0
所以 k = 2
若α = β，則
方程式有重根
B2 – 4AC = 0
[-(k-2)]2 – 4(1)[-(k -1)] = 0
k2 – 4k + 4 + 4k - 4 = 0
k2 = 0
k = 0
2006-10-17 4:18 am
α, β are the real roots of the equation x^2+ (k-2)x-(k-1)=0.
Because it |α | , | β | are absoulte numbers, so we need to do it with cases.
Case 1, if α= -β
α + β = 0
(k-2)=0
k=2
Case 2, if α=β
B^2-4AC=0
(k-2)^2-4(1)[-(k-1)]=0
k^2-4k+4+4k-4=0
k^2=0
k=0
From the result of Case 1 and 2, we know k=2 or k=0
Because it |α | , | β | are absoulte numbers, so we need to do it with cases.
Case 1, if α= -β
α + β = 0
(k-2)=0
k=2
Case 2, if α=β
B^2-4AC=0
(k-2)^2-4(1)[-(k-1)]=0
k^2-4k+4+4k-4=0
k^2=0
k=0
From the result of Case 1 and 2, we know k=2 or k=0
參考: me
2006-10-17 4:09 am
| α | = | β |,
α =β / α=-β
α -β=0/α +β=0
=============
α +β=2-k
αβ=1-k
α +β=0
2-k=0
k=2
α -β=0
[(2-k)^2-4(1-k)]^1/2=0
(4-4k+k^2-4+4k)^1/2=0
k^2=0
k=0
Therefore,k=0 or k=2
α =β / α=-β
α -β=0/α +β=0
=============
α +β=2-k
αβ=1-k
α +β=0
2-k=0
k=2
α -β=0
[(2-k)^2-4(1-k)]^1/2=0
(4-4k+k^2-4+4k)^1/2=0
k^2=0
k=0
Therefore,k=0 or k=2
2006-10-17 4:07 am
| α | = | β |
α=β or α=-β
α+β=0
-(k-2)=0
k=2
since that α=β,so △=0
(k-2)^2-4(1)(-k+1)=0
k^2-4k+4+4k-4=0
k^2=0
k=0
so k=0 or 2
α=β or α=-β
α+β=0
-(k-2)=0
k=2
since that α=β,so △=0
(k-2)^2-4(1)(-k+1)=0
k^2-4k+4+4k-4=0
k^2=0
k=0
so k=0 or 2
收錄日期: 2021-04-22 00:13:36
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061016000051KK03481