F.4 a.maths

2007-10-22 5:13 am
1. 若α和β為方程3x^2 + 6x + 7 = 0的根 , 試求下列各式之值:
a) α^2 β + β^2 α

b) α^4 + β^4

回答 (1)

2007-10-22 5:37 am
✔ 最佳答案
a)
α + β = -(6)/3 = -2
α β = 7/3

α^2 β + β^2 α
= α β (α + β) <--- 因式分解
= (7/3) * (-2)
= -14/3

b)
α^4 + β^4
= (α^4 + β^4 + 2α^2β^2 -2α^2β^2)
= (α^2 + β^2)^2 -2α^2β^2
= (α^2 + β^2 + 2αβ - 2αβ)^2 -2α^2β^2
= [(α+β)^2 -2αβ]^2 -2α^2β^2
= [(-2)^2 - 2(7/3)]^2 -2(7/3)^2
= [4 - (14/3)]^2 -2(49/9)
= (-2/3)^2 - 98/9
= 4/9 - 98/9
= -94/9

有錯請指正!
參考: My Maths knowledge


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