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#1 2006-12-28 09:26:10
- Neha
- Member
- Registered: 2006-10-11
- Posts: 173
question
Apply Descartes' Rule of Signs to determine the possible numbers of positive roots of:
x^4 - x^3 - 3x^2 - x + 2 = 0
but not sure what i did is right....correct me if i did it wrong
f(x) = +x^4 - x^3 - 3x^2 - x + 2
f(-x) = -x^4 - (-x)^3 - 3(-x)^2 - (-x) + 2
= -x^4 + x^3 + 3x^2 + x + 2
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#2 2006-12-30 04:20:43
- er.neerajsrivastava
- Member
- Registered: 2006-12-27
- Posts: 9
Re: question
no of possitive roots = the no of times sign is changed in the eq f(x)
and no of negative roots = the no of times sign changes in eq f(-x).
so for the given eq. +ve root = 2
and -ve root =
f(-x)= x^4 + x^3 -3x^2 +x +2=0
so -ve root = 2
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#3 2006-12-30 04:31:45
- krassi_holmz
- Real Member
- Registered: 2005-12-02
- Posts: 1,905
Re: question
Hm, Interesting, I didn't know that before.
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