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 #作者: scotten 時間: 06-3-19 00:17 標題: 回覆: 代數的3個問題 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 #作者: yacool5210 時間: 06-3-19 00:25 標題: 回覆: 代數的3個問題 [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作者: yacool5210 時間: 06-3-19 00:25 標題: 回覆: 代數的3個問題 [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]