中四附加數！！緊急！(想了很久我也不懂做)

1a)
α+β=k+2;αβ=k

1b)
(α+1)(β+2)=4
αβ+2α+β+2=4
αβ+α+(α+β)=2
(k)+α+(k+2)=2
...............α=-2k

2)
因為α、β為二次方程x^2+px+q=0的根,所以
α+β=-p..............(1)
αβ=q.................(2)
因為α+3、β+3為方程x^2+qx+p=0的根,所以
(α+3)+(β+3)=-q
α+β+6=-q
(-p)+6=-q
p-q=6.............(3)

(α+3)(β+3)=p
αβ+3α+3β+9=p
(q)+3(-p)+9=p
4p-q=9............(4)

(4)-(3) 3p=3
...........p=1
...........q=-5

3)
a)α+β=2-p;αβ=p

b)
α^2+β^2=11
α^2+β^2+2αβ-2αβ=11
(α+β)^2-2αβ=11
(2-p)^2-2p=11
p^2-6p-7=0
(p+1)(p-7)=0
p=-1 or p=7

當 p=-1,
x^2+(p-2)x+p=x^2-3x-1=0
Δ =3^2-4(1)(-1)=13>0

當 p=7,
x^2+(p-2)x+p=x^2+5x+7=0
Δ =5^2-4(1)(7)=-24<0........ 因為 α及β為實數,所以p不可以等於7
所以, p=-1

2007-10-20 02:38:44 補充：
1b)補充:因為α=-2k為二次方程x^2-(k+2)x+k=0的根,所以α^2-(k+2)α+k=0(-2k)^2+2k(k+2)+k=06k^2+5k=0k(6k+5)=0k=0 or k=-5/6

1.
(a)
方程中的a=1, b= - (k+2), c=k
α+β = -b/a = -[-(k+2)/1] = k+2
αβ = c/a = k/1 = k
(b)
(α+1)(β+2)=4
αβ+2α+β+2=4
αβ+α+(α+β)+2=4
(k)+α+(k+2)+2=4 <-- 代入由(a)得到的有關數字
α+2k=0
α= -2k
將 α= -2k 代入 α+β = k+2
-2k+ β = k+2
β = 3k+2
將 α= -2k 和 β = 3k+2 代入 αβ = k
(-2k)(3k+2)=k
-6k^2-4k=k
6k^2+5k=0
k(6k+5)=0
k = 0 或k = -5/6

2.
方程中的a=1, b=p, c=q
α+β = -b/a = -(p/1) = -p
αβ = c/a = q/1 = q

(α+3)+(β+3) = α+β+6 = -p+6 = -q
(α+3)(β+3) = αβ+3α+3β+9 = αβ+3(α+β)+9 = q-3p+9 = p
得出兩條二元一次方程:
p-q = 6 ...(1)
4p-q = 9 ...(2)
(2) - (1): 3p = 3 --> p = 1
將 (p = -5) 代入 (1),
1-q = 6 --> q = -5

3.
(a)
方程中的a=1, b=p-2, c=p
α+β = -b/a = -[(p-2)/1] = -p+2 = 2-p
αβ = c/a = p/1 = p
(b)
α^2+β^2=11
α^2+β^2+2αβ-2αβ=11
(α+β)^2-2αβ=11 <-- [因為α^2+β^2+2αβ = α^2+2αβ+β^2]
(2-p)^2-2p=11
4-4p+p^2-2p=11
p^2-6p-7=0
(p-7)(p+1)=0
p=7 或 p= -1
當p=7, 判別式=(7-2)^2-4(7)<0. 無實根.
當p= -1, 判別式=(-1-2)^2-4(-1)>0. 有實根.
所以, p= -1

有錯請指正!

幫樓上補充一下漏左的1(b)後面部份~

α = -2k
since α + β = k + 2
β = k + 2 + 2k
β = 3k + 2

(α + 1)(β + 2)=4
(-2k + 1)(3k + 2 + 2)=4
(1 - 2k)(3k + 4)=4
3k + 4 -6k^2 -8k =4
6k^2 + 5k =0
k(6k + 5)=0
k=0 or -5/6

2007-10-20 02:41:34 補充：
原來補左~唔使理我~飛走~