Quadratic Equations
2014-10-24 7:45 pm
If α and 1/α are the roots of the equation 3x^2 + (k^2 + 1)x + k = 0, find thevalues ofa. k (I calculated)b.α + 1/α(I calculated)c.α^2 + 1/α^2 (I do not know how to do)d. α^3 + 1/α^3(I do not know how to do)
回答 (2)
2014-10-24 8:42 pm
✔ 最佳答案
a) As α and 1/α are the roots of the equation 3x² + (k² + 1)x + k = 0,so, α + 1/α = -(k² + 1)/3 and α * 1/α＝k/3, that is,1 = k/3∴ k＝3b) α + 1/α＝-(k² + 1)/3＝-10/3
c) α² + 1/α²＝α² + 2 + 1/α² - 2＝(α + 1/α)² - 2＝100/9 - 2 ⋯⋯⋯⋯⋯⋯ (α + 1/α＝-10/3)＝82/9
d) α³ + 1/α³＝(α + 1/α)(α² - 1 + 1/α²)＝(-10/3)(82/9 - 1) ⋯⋯⋯⋯ (α + 1/α＝-10/3, α² + 1/α²＝82/9) ＝-730/27
2014-10-25 1:07 am
c.α^2 + 1/α^2
(α + 1/α)^2
= α^2 + 2 + 1/α^2
= (α^2 + 1/α^2) + 2
Hence α^2 + 1/α^2 = (α + 1/α)^2 - 2
The answer can be obtained simply by putting in the result of b for (α + 1/α).
d. α^3 + 1/α^3
(α + 1/α)^3
= (α^2 + 2 + 1/α^2) x (α + 1/α)
= α^3 + 2α + 1/α + α + 2/α + 1/α^3
= α^3 + 3α + 3/α + 1/α^3
= (α^3 + 1/α^3) + 3(α + 1/α)
Hence α^3 + 1/α^3 = (α + 1/α)^3 - 3(α + 1/α)
The answer can be obtained simply by putting in the result of b for (α + 1/α).
(α + 1/α)^2
= α^2 + 2 + 1/α^2
= (α^2 + 1/α^2) + 2
Hence α^2 + 1/α^2 = (α + 1/α)^2 - 2
The answer can be obtained simply by putting in the result of b for (α + 1/α).
d. α^3 + 1/α^3
(α + 1/α)^3
= (α^2 + 2 + 1/α^2) x (α + 1/α)
= α^3 + 2α + 1/α + α + 2/α + 1/α^3
= α^3 + 3α + 3/α + 1/α^3
= (α^3 + 1/α^3) + 3(α + 1/α)
Hence α^3 + 1/α^3 = (α + 1/α)^3 - 3(α + 1/α)
The answer can be obtained simply by putting in the result of b for (α + 1/α).
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