Let α and β be root of equation x^2+px-2p^2=0
(a)find a quadratic equation which roots are (α^2-β) and (β^2-α)
also,express α^3+β^3
f.4 amath
2007-11-13 4:29 am
回答 (2)
2007-11-13 4:40 am
✔ 最佳答案
As follows~~~圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/ScreenHunter_03Nov122039.jpg?t=1194871235
參考: Myself~~~
2007-11-13 4:49 am
(α^2-β) +(β^2-α)
=(α+β)^2 - 2αβ-(α+β)
=p^2+4p^2+p
=5p^2+p
(α^2-β) (β^2-α)
= (αβ)^2- (α^3+β^3)+ αβ
=(αβ)^2- (α+β)[(α+β)^2-3αβ]+αβ
=4p^4-(-p)(p^2+6p^2)-2p^2
=4p^4+7p^3-2p^2
equation:x^2-(5p^2+p)x+(4p^4+7p^3-2p^2)=0
α^3+β^3=(α+β)[(α+β)^2-3αβ]
=(-p)(p^2+6p^2)
=-7p^3
=(α+β)^2 - 2αβ-(α+β)
=p^2+4p^2+p
=5p^2+p
(α^2-β) (β^2-α)
= (αβ)^2- (α^3+β^3)+ αβ
=(αβ)^2- (α+β)[(α+β)^2-3αβ]+αβ
=4p^4-(-p)(p^2+6p^2)-2p^2
=4p^4+7p^3-2p^2
equation:x^2-(5p^2+p)x+(4p^4+7p^3-2p^2)=0
α^3+β^3=(α+β)[(α+β)^2-3αβ]
=(-p)(p^2+6p^2)
=-7p^3
收錄日期: 2021-04-29 19:44:31
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