標題:
Maths
If a, α are the roots of the equation x^2 + 16x+ 69 =0 and a,β are the roots of the given equation x^2 -2x -35=0 ,without solving the two given equations,find the values of a) β -α , b) α / β
最佳解答:
a+ α = -16 and aα = 69 a + β = 2 and aβ = -35 (a) Consider a+ α = -16...(1) and a + β = 2 ...(2) (2) - (1) a+β - (α+a) = 2 - (-16) β -α = 18 (b) Consider aα = 69 ...(3) and aβ = -35 ...(4) (3)/(4) : α / β = 69/(-35) = -69/35
其他解答:
a+ α = -16 and aα = 69 a + β = 2 and aβ = -35 (a) Consider a+ α = -16...(1) and a + β = 2 ...(2) (2) - (1) a+β - (α+a) = 2 - (-16) β -α = 18 (b) Consider aα = 69 ...(3) and aβ = -35 ...(4) (3)/(4) : α / β = 69/(-35) = -69/35|||||a + α = -16 a x α = 69 a + β = 2 a x β = -35 so,, β -α = (a + β) - (a + α) = 2 - (-16) = 18 also,, α / β = (a x β) / (a x α) = -35 / 69 2007-10-28 15:59:09 補充： 錯左@@also,, α / β = (a x β) / (a x α) = -35 / 69應該係(a x α) / (a x β) = -69/35 先岩@@
Maths
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發問:If a, α are the roots of the equation x^2 + 16x+ 69 =0 and a,β are the roots of the given equation x^2 -2x -35=0 ,without solving the two given equations,find the values of a) β -α , b) α / β
最佳解答:
a+ α = -16 and aα = 69 a + β = 2 and aβ = -35 (a) Consider a+ α = -16...(1) and a + β = 2 ...(2) (2) - (1) a+β - (α+a) = 2 - (-16) β -α = 18 (b) Consider aα = 69 ...(3) and aβ = -35 ...(4) (3)/(4) : α / β = 69/(-35) = -69/35
其他解答:
a+ α = -16 and aα = 69 a + β = 2 and aβ = -35 (a) Consider a+ α = -16...(1) and a + β = 2 ...(2) (2) - (1) a+β - (α+a) = 2 - (-16) β -α = 18 (b) Consider aα = 69 ...(3) and aβ = -35 ...(4) (3)/(4) : α / β = 69/(-35) = -69/35|||||a + α = -16 a x α = 69 a + β = 2 a x β = -35 so,, β -α = (a + β) - (a + α) = 2 - (-16) = 18 also,, α / β = (a x β) / (a x α) = -35 / 69 2007-10-28 15:59:09 補充： 錯左@@also,, α / β = (a x β) / (a x α) = -35 / 69應該係(a x α) / (a x β) = -69/35 先岩@@
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