代數的3個問題

yacool5210

1.設 i = √-1 a屬於R 若X的方程式 X^2-(a+i)X+(-2+2i)=0 有實根 則a=??


2.若X^2+(a-3)X+a+6=0之兩根均為整數 求a=??


3.設α β 為X^2+X+1=0 之兩根 則以 (1+α^5)/(1+β^5) , (1+β^5)/(1+α^5)為兩根方程式為??

yacool5210

[quote=scotten]1.
x^2-(a+i)x+(2i-2)=0
=>(x^2-ax-2)+i(2-x)=0
because a,x are reals,(x^2-ax-2) and (2-x) are reals
=>x^2-ax-2=0 and 2-x=0
=>x=2 and a=1 #[/quote]

恭喜你 這題也答對了~~

yacool5210

[quote=scotten]2.
x^2+(a-3)x+(a+6)=0

let two roots be s, t (s>t)
then s+t=3-a,st=a+6
(s+1)(t+1)=st+s+t+1=(a+6)+(3-a)+1=10
becase s,t are integers ,s+1,t+1 are also integers.

s+1=10,5,-1 ,-2
t+1=1 ,2,-10,-5
=>(s,t)=(9,0),(4,1),(-2,-11),(-3,-6)
=> a=-6,-2,16 or 12 #[/quote]

YES
your answer is true

scotten

1.
x^2-(a+i)x+(2i-2)=0
=>(x^2-ax-2)+i(2-x)=0
because a,x are reals,(x^2-ax-2) and (2-x) are reals
=>x^2-ax-2=0 and 2-x=0
=>x=2 and a=1  #

scotten

2.
x^2+(a-3)x+(a+6)=0

let two roots be s, t (s>t)
then s+t=3-a,st=a+6
(s+1)(t+1)=st+s+t+1=(a+6)+(3-a)+1=10
becase s,t are integers ,s+1,t+1 are also integers.

s+1=10,5,-1 ,-2
t+1=1 ,2,-10,-5
=>(s,t)=(9,0),(4,1),(-2,-11),(-3,-6)
=> a=-6,-2,16 or 12  #

yacool5210

[quote=M.N.M.]3.設α β 為X^2+X+1=0 之兩根 則以 (1+α^5)/(1+β^5)，(1+β^5)/(1+α^5)為兩根方程式為??
囧，發完後要檢查題目有無打錯的地方
α+β=-1
αβ=1

α^2+α+1=0
=>α^3=1
=>α^5=α^2=-α-1
同理β^5=β^2=-β-1

(1+α^5)/(1+β^5)
=(1+α^2)/(1+β^2)
=α/β
同理(1+β^5)/(1+α^5)=β/α

α/β+β/α=(α^2+β^2)/αβ=[(α+β)^2-2αβ]/αβ=-1
α/β*β/α=1

所以 (1+α^5)/(1+β^5)，(1+β^5)/(1+α...[/quote]


答對嚕~~~
打太快沒注意  下次會改進!!

M.N.M.

3.設α β 為X^2+X+1=0 之兩根 則以 (1+α^5)/(1+β^5)，(1+β^5)/(1+α^5)為兩根方程式為??
囧，發完後要檢查題目有無打錯的地方
α+β=-1
αβ=1

α^2+α+1=0
=>α^3=1
=>α^5=α^2=-α-1
同理β^5=β^2=-β-1

(1+α^5)/(1+β^5)
=(1+α^2)/(1+β^2)
=α/β
同理(1+β^5)/(1+α^5)=β/α

α/β+β/α=(α^2+β^2)/αβ=[(α+β)^2-2αβ]/αβ=-1
α/β*β/α=1

所以 (1+α^5)/(1+β^5)，(1+β^5)/(1+α^5)為兩根方程式為x^2+x+1=0