F4 Nature of roots

2011-10-16 6:54 am
Q2 Given that r is a root of the quadratic equationx^2-3x-5=0, find the valueof r^5-3r^4-5r^3+2r^-6r

回答 (1)

2011-10-16 7:06 am
✔ 最佳答案
Since r is a root of the quadratic equation x^2 - 3x - 5 = 0,
i.e. r^2 - 3r - 5 = 0
r^5-3r^4-5r^3+2r^2-6r
= r^3 (r^2 - 3r - 5) + 2r^2 - 6r - 10 + 10
= r^3 (r^2 - 3r - 5) + 2(r^2 - 3r - 5) + 10
= 0 + 0 + 10
= 10
參考: Hope the solution can help you^^”


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