『A.Math 』有關Quadratic formula ---α同β

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|
α=β
α+β = -(k-2)
2α = -(k-2)
α = -(k-2) /2 -----(1)
α x β = - (k -1)
α^2 = - (k -1) ------(2)

put (1)into (2)

[ -(k-2) /2]^2 = - (k -1)
k^2-4k+4/4 = -k+1 -----將左面的4移過右面
k^2-4k+4 = -4k+4
k^2-4k+4k+4-4=0
k^2=0
k =0

要計埋α=-β

α=-β
α+(-β) = - ( k-2 )
α - α = -( k-2 )
0 = -k +2
-2 = -k
k=2

α x (-β) = - (k -1)

-α^2 = - (k -1)
α^2 = (k -1)
α = (k -1)的開方

由於上面α x (-β) = - (k -1)
計唔到k 呢個就可捨去
所以k=0 ,2

X^2 + ( k-2 ) x - (k -1) = 0

a+b= -(k-2)
ab=-(k-1) a,b= roots

|a| = |b|
(a)^2=(b)^2
(a)^2-(b)^2=0
(a+b)(a-b)=0
(a+b) roots(a-b)^2=0 roots=平方根
(a+b) roots(a^2-2ab+b^2) =0
(a+b) roots[(a+b)^2-4ab]=0

所以 [-( k-2 )] roots[( k-2 )^2+4(k -1)]=0
[-( k-2 )] roots[k^2-4k+4+4k-4]=0
[-( k-2 )] k=0
k^2-4k=0
k=0 , 2

b^2-4ac=0, since they are the same point
(k-2)^2+4(k-1)=0
k^2-4k+4+4k-4=0
k=0

so the points are -1 0r 1
(x-1)(x+1)=x^2-1
using comparison with x ^2 + ( k-2 ) x - (k -1)
k=2

首先，您要 factorize 原本條 formula，搵出 2個 roots，步驟如下：

x ^2 + ( k-2 ) x - (k -1) = 0
　[ x+ ( k-1 ) ] [ x -1 ] = 0

thus, the roots are k-1 & -1

=> | k-1 | = | -1 |

=> k-1 = 1 or -( k-1 ) = 1

=> k = 2 or k = 0

希望幫到您啦~ ^^

x^2 + ( k-2 ) x - (k -1) = 0

|α| = |β| ,

There are two cases:
α = β or α = -β

Case α = β
α + β = -(k-2)
2α = -(k-2)
α = -1/2(k-2)

αβ = -(k-1)
α * α = -(k-1)
-1/2(k-2) * -1/2(k-2) = -(k-1)
1/4(k-2)^2 = -k + 1
k^2 - 4k + 4 = -4k + 4
k^2 = 0
k = 0

Case α = -β
α + β = -(k-2)
0 = -(k-2)
0 = k -2
k = 2

So, we get k = 0 or k = 2