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文章數：239 |
 F4 A-MATHS 急...... |
| 休閒生活｜影視戲劇 2017/08/09 00:01:15 |
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此文章來自奇摩知識+如有不便請留言告知 其他解答:Here is a method WITHOUT DIFFERENTIATION: r^3 + (2-α)r^2 + (β-2α+1)r - (α+1) = 0 r =2 is a double root. (r-2)^2 is a factor of r^3 + (2-α)r^2 + (β-2α+1)r - (α+1) Therefore, r^3 + (2-α)r^2 + (β-2α+1)r - (α+1) = (r-2)2(r-c) r^3 + (2-α)r^2 + (β-2α+1)r - (α+1) = r^3 - (c+4)r^2 + (4c+4)r - 4c Compare the r^2 term: 2-α = -(c+4) …… (1) Compare the r term: β-2α+1 = 4c+4 …… (2) Compare the constant term: -(α+1) = -4c …… (3) (1): 2-α = -c-4 c =α-6 …… (4) 4c = 4α-24…… (5) (3) 4c =α+1 ….. (6) (5)=(6) 4α-24 = α+1 3α = 25 α = 25/3 Put α=25/3 into (4) c = (25/3)-6 c = 7/3 Put α=25/3 and c=7/3 into (2): β-2(25/3)+1 = 4(7/3)+4 β-47/3 = 40/3 β = 29 Ans: α = 25/3, β = 29|||||Cubic equations should be something in pure maths rather than amaths... |
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