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Descartes' Rule of Signs and Complex Roots


Date: 08/04/98 at 02:02:17
From: Ty McDowell
Subject: Descartes' Rule of Signs

I am having an extremely difficult time understanding what I am 
supposed to do here. The question is prove that if p and q are 
real and q is not equal to 0, the equation x^3 + px + q = 0 has two 
imaginary roots. Can you help me on this one?


Date: 08/04/98 at 16:51:03
From: Doctor Rob
Subject: Re: Descartes' Rule of Signs

Use Descartes' Rule of Signs twelve times:

If p > 0, q > 0, then there are no positive real roots, and by looking
at -x^3 - p*x + q = 0, the original equation has at most one negative
real root, so there must by two complex roots.

If p > 0, q < 0, then there is at most one positive real root, and by
looking at -x^3 - p*x + q = 0, the original equation has no negative
real roots, so there must be two complex roots.

If p < 0, q > 0, then there are at most two positive real roots, and 
by looking at -x^3 - p*x + q = 0, the original equation has at most 
one negative real root. This gives us enough information to construct 
a counterexample: the sum of the roots must be zero since the 
coefficient of x^2 is zero, so (x-2)*(x-1)*(x+3) = x^3 - 7*x + 6 has 
no complex roots.

If p < 0, q < 0, then there is at most one positive real root, and by
looking at -x^3 - p*x + q = 0, the original equation has at most two
negative real roots. This gives us enough information to construct a
counterexample: the sum of the roots must be zero since the 
coefficient of x^2 is zero, so (x+2)*(x+1)*(x-3) = x^3 - 7*x - 6 has 
no complex roots.

If p = 0, q > 0, then there is no positive real root and at most one
negative real root, so there must be two complex roots.

If p = 0, q < 0, then there is at most one positive real root and no
negative real root, so there must be two complex roots.

The actual condition is that the equation has two complex roots if and 
only if -(q/2)^2 < (p/3)^3. The proof of this statement is somewhat 
involved, and not the point of the question, but I thought you would 
be interested in knowing the facts.

- Doctor Rob, The Math Forum
Check out our web site! http://mathforum.org/dr.math/   
    
Associated Topics:
High School Analysis
High School Polynomials

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