Descartes' Rule of Signs and Complex RootsDate: 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/ |
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