If the roots of x2-x+1=0 are α and β, create a quadratic equation that has the roots 1÷(α+1) and 1÷(β+1).
*(Note that the coefficent of x2 is 1.)
Root-Coefficent Relationships(2)
2007-11-11 9:07 pm
回答 (2)
2007-11-11 9:14 pm
✔ 最佳答案
α + β = 1, αβ = 11/(α+1) + 1/(β+1) = (α+1+β+1)/(α+1)(β+1)
= (1+2)/(αβ+α + β+1)
= 3/(1+1+1)
= 1
1/(α+1) * 1/(β+1)
= 1/(α+1)(β+1)
= 1/(αβ+α + β+1)
= 1/(1+1+1)
= 1/3
The equation is x^2 - x + 1/3 = 0
3x^2 - 3x + 1 = 0
2007-11-11 9:17 pm
α+β = -1 / 1 = -1
αβ = 1 / 1 = 1
[1/(α+1)] + [1/(β+1)]
=(α+β+2) / (αβ+α+β+1)
=(1+2) / (1+1+1)
=1
[1/(α+1)] [1/(β+1)]
=1 / (αβ+α+β+1)
=1 / 3
Therefore, the equation having the roots 1÷(α+1) and 1÷(β+1) is
x^2-x+(1/3)=0
(So sum of roots = 1, product of roots = 1/3)
αβ = 1 / 1 = 1
[1/(α+1)] + [1/(β+1)]
=(α+β+2) / (αβ+α+β+1)
=(1+2) / (1+1+1)
=1
[1/(α+1)] [1/(β+1)]
=1 / (αβ+α+β+1)
=1 / 3
Therefore, the equation having the roots 1÷(α+1) and 1÷(β+1) is
x^2-x+(1/3)=0
(So sum of roots = 1, product of roots = 1/3)
收錄日期: 2021-04-13 18:12:08
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https://hk.answers.yahoo.com/question/index?qid=20071111000051KK01697