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ContentTheory of Equation
Theory of EquationDescartes' Rule of Signs416 In the following theorems every two adjacent terms in π(π₯), which have the same signs, count as one "continuation of sign"; and every two adjacent terms, with different signs, count as one change of sign. 417 π(π₯), multiplied by (π₯βπ), has an odd number of changes of sign thereby introduced, and one at least. 418 π(π₯) cannot have more positive roots than changes of sign, or more negative roots than continuations of sign. 419 When all the roots of π(π₯) are real, the number of positive roots is equal to the number of changes of sign in π(π₯); and the number of negative roots is equal to the number of changes of sign in π(βπ₯). 420 Thus, it being known that the roots of the equation π₯4β10π₯3+35π₯2β50π₯+24=0 are all real; the number of positive roots will be equal to the number of changes of sign, which is four. Also π(βπ₯)=π₯4+10π₯3+35π₯2+50π₯+24=0, and since there is no change of sign, there is consequently, by the rule, no negative root. 421 If the degree of π(π₯) exceeds the number of changes of sign in π(π₯) and π(βπ₯) together, by π, there are at least π imaginary roots. 422 If, between two terms in π(π₯) of the same sign, there be an odd number of consecutive terms wanting, then there must be at least one more than that number of imaginary roots; and if the missing terms lie between terms of different sign, there is at elast one less than the same number of imaginary roots. Thus, in the cubic π₯3+4π₯β7=0< There must be two imaginary roots. And in the equation π₯6β1=0< there are, for certain, four imaginary roots. 423 If an even number of consecutive terms be wanting in π(π₯), there is at least the same number of imaginary roots. Thus the equation π₯5+1=0 has four terms absent; and therefore four imaginary roots at least.Sources and Referenceshttps://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdriveΒ©sideway ID: 210800005 Last Updated: 8/5/2021 Revision: 0 Ref: References
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