1)if αand β are the root of the equation 3x2(2代表2次方)+6x +7,find the values of
α4+ β4(4代表4次方)
2)if m and n are the roots of the equation x2(2代表2次方)+px + q =0,express the following in terms of
(a) (1-m)(1-n) (b)m2(2代表二次方)/n+ n2(2代表二次方)/m
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2007-10-01 7:04 pm
回答 (3)
2007-10-01 7:16 pm
✔ 最佳答案
1. α+ β = -2 and αβ = 7/3α^4 + β^4 = (α^2+ β^2)^2 -2α^2β^2
=[(α+ β)^2-2αβ]^2 - 2α^2β^2
=[(-2)^2 -14/3)^2 -2(7/3)^2
=4/9-98/9
=-94/9
2. m+n =-p and mn = q
(a) (1-m)(1-n) = 1-(m+n) +mn = 1+p+q
(b) m^2/n+ n^2/m = (m^3+n^3)/mn
=(m+n)(m^2-mn+n^2)/mn
=(m+n)[(m+n)^2-3mn]/mn
=(-p)[(-p)^2-3q]/q
=p(3q-p^2)/q
2007-10-01 7:22 pm
1)α+β=-2,αβ=7/6
α^4+β^4=(α^2+β^2)^2-2(αβ)^2
............=((α+β)^2-2αβ)-2(αβ)^2
............=-19/18
2)m+n=-p,mn=q
唔知要做咩
in terms of 咩?
點解會有個b?
2007-10-01 11:35:38 補充：
哦原來係分左2part(a)1減(m加n)加mn=1加p加q(b)(m^3加n^3)/mn=(m加n)(m^2減mn加n^2)/mn=p(2q減p^2)/q
α^4+β^4=(α^2+β^2)^2-2(αβ)^2
............=((α+β)^2-2αβ)-2(αβ)^2
............=-19/18
2)m+n=-p,mn=q
唔知要做咩
in terms of 咩?
點解會有個b?
2007-10-01 11:35:38 補充：
哦原來係分左2part(a)1減(m加n)加mn=1加p加q(b)(m^3加n^3)/mn=(m加n)(m^2減mn加n^2)/mn=p(2q減p^2)/q
2007-10-01 7:20 pm
sum of root = -6/3 = -2, product of root = 7/3
1) α4+ β4 = α^4+ β^4 + 2α^2β^2 - 2α^2β^2 = (α^2+β^2)^2 - 2α^2β^2 跟住再拆就搞掂。
2) a) 1- (m+n) + mn
b) (m^3+n^3)/mn = (m+n) (m^2-mn+n^2) / mn 跟住一樣做
1) α4+ β4 = α^4+ β^4 + 2α^2β^2 - 2α^2β^2 = (α^2+β^2)^2 - 2α^2β^2 跟住再拆就搞掂。
2) a) 1- (m+n) + mn
b) (m^3+n^3)/mn = (m+n) (m^2-mn+n^2) / mn 跟住一樣做
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