A MATH

α and β are the roots>>> (x-α)(x-β)=0
x^2-(α+β)x+αβ=0
α^2 and β^2 are the roots>>> (x-α^2)(x-β^2)=0
x^2-(α^2+β^2)x+(αβ)^2=0

-(α+β)=p
-αβ=6
(α^2+β^2)=13
(αβ)^2=q

(αβ)^2=q
q=36~~
(α^2+β^2)+2αβ=13-2x6
(α+β)^2=1
α+β= -/+1
-p= -/+1
p= -/+1~~
checking
by sub. α= -2 and β= 3
or α= -3 and β= 2
or α= 2 and β= -3
or α= 3 and β= -2
right@@

2007-09-11 19:19:30 補充：
α β = -p/1?????????? 補充時間：2007-09-11 17:50:40how come the -6?????? 補充時間：2007-09-11 17:53:57how come the -6?????? 補充時間：2007-09-11 17:53:58以上個d野係用 你由roots 整出黎個條式 同你題目個條做對比 由於由roots整出黎個條式同你題目個條一定係同一條式 so 你個d 答案就會出哂黎 明未???

α+β=-p αβ=-6

α^2+β^2=13 α^2β^2=q

(α+β)^2-2αβ=13 (αβ)^2=α^2β^2
p^2+12=13 q=36
p=+-1