Explore BrainMass

Explore BrainMass

    Roots of Polynomial for a Derivative

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    An elliptic curve can be written as y^2=x^3+ax+b. I need a proof for why x^3+ax+b either have 3 real roots or 1 real root and 2 complex roots. I don't have anything that I know about it prior to asking for help here at Brainmass.

    © BrainMass Inc. brainmass.com October 2, 2022, 12:45 pm ad1c9bdddf
    https://brainmass.com/math/calculus-and-analysis/roots-polynomial-derivative-630061

    SOLUTION This solution is FREE courtesy of BrainMass!

    See attachment

    Problem:

    An elliptic curve can be written as y^2=x^3+ax+b. I need a proof for why x^3+ax+b either have 3 real roots or 1 real root and 2 complex roots.

    Solution:

    It is important first to point out that the coefficients a and b are real, otherwise the given polynomial will have only complex roots. In this case, one needs to analyze the polynomial function
    where ( 1)
    One first remark: ( 2)
    Since f(x) is a continuous function, it means that there exists at least one point for which . In other words, the equation admits always at least one real solution.
    The problem now is te nature of the other two solutions of the equation . The following cases are possible:

    1) The derivative of f(x)
    ( 3)
    has real roots. In this case, it is necessary that . Let's denote the (real) solutions of equation :
    ( 4)
    If the values of f(x) in those points are of opposite sign:
    ( 5)
    then, the equation will have 3 real solutions.
    Otherwise, the equation will have one real solution and two complex solutions.
    For the particular case when or , the equation will have 3 real solutions, but one of them will be double ( ).
    2) The derivative of f(x)
    ( 6)
    has complex roots. This happens when . In this case, the function f(x) will be monotonic on R (increasing, there are no extrema) and, as a consequence, the equation will have only one real solution.
    Let's suppose now that one root of equation is complex. According to one of Viete formulas, we have
    ( 7)
    Since and , it follows that . This happens only when and are complex conjugate:
    ( 8)

    As such, it is not possible for the given polynomial with real coefficients to have only one complex root: it will have either two complex roots (which will be necessarily conjugate), or only real roots.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com October 2, 2022, 12:45 pm ad1c9bdddf>
    https://brainmass.com/math/calculus-and-analysis/roots-polynomial-derivative-630061

    Attachments

    ADVERTISEMENT