quadratic equation

2009-09-25 2:58 am

Given that r is a root of the quadratic equation x^2-3x-5=0, find the value of r^5-3r^4-5r^3+2r^2-6r.

THANK U!

回答 (2)

2009-09-25 3:34 am

✔ 最佳答案

r^5 - 3r^4 - 5r^3 + 2r^2 - 6r
= r^3(r^2 - 3r - 5) + 2r^2 - 6r
= r^3 (0) + 2r^2 - 6r
= 2(r^2 - 3r)
= 2 * 5
= 10

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2009-09-25 3:37 am

r^5-3r^4-5r^3+2r^2-6r
=r(r^4-3r^3-5r^2+2r-6)
=r[r^2(r^2-3r-5)+2r-6]
=r(2r-6)
=2r(r-3)
Since r^2-3r=5=>r-3=5/r
So 2r(r-3)=2r(5/r)=10

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