quadratic equation

α 與 β 是real roots of the eqation
x2 + ( 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

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

| 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 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