a-maths
回答 (2)
2007-07-04 5:30 am
✔ 最佳答案
Since α , β are the roots of x² - 3x + p = 0So, product of roots, αβ = p
Sum of roots, α + β = 3
α^4 + β^4 = 17
(α² + β²)² - 2α²β² = 17
[(α + β)² - 2αβ]² - 2(αβ)² = 17
[(3)² - 2p]² - 2p² = 17
81 - 36p + 4p² - 2p² = 17
p² - 18p + 32 = 0
(p - 16)(p - 2) = 0
p = 2 or 16
參考: Myself~~~
2007-07-04 5:36 am
consider α+β=3
αβ=p
(α+β)^4=α^4 +4α^3β +6α^2β^2 +4αβ^3 +β^4
α^4+β^4=(α+β)^4 -4α^3β -6(αβ)^2 -4αβ^3
=(α+β)^4 -6(αβ)^2 -4αβ(α^2+β^2)
=(α+β)^4 -6(αβ)^2 -4αβ[(α+β)^2-2αβ]
=81-6p^2 -4p(9-2p)
=2p^2-36p+81
α^4+β^4 =17
2p^2-36p+81=17
2p^2-36p+64=0
p=16 or2
αβ=p
(α+β)^4=α^4 +4α^3β +6α^2β^2 +4αβ^3 +β^4
α^4+β^4=(α+β)^4 -4α^3β -6(αβ)^2 -4αβ^3
=(α+β)^4 -6(αβ)^2 -4αβ(α^2+β^2)
=(α+β)^4 -6(αβ)^2 -4αβ[(α+β)^2-2αβ]
=81-6p^2 -4p(9-2p)
=2p^2-36p+81
α^4+β^4 =17
2p^2-36p+81=17
2p^2-36p+64=0
p=16 or2
收錄日期: 2021-04-23 20:33:55
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https://hk.answers.yahoo.com/question/index?qid=20070703000051KK04410