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Thread: Discriminant question

  1. #1
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    Discriminant question

    Using the Theorem on Symmetric Functions or otherwise,
    show that the discriminant of the quartic polynomial x^4 + qx + r
    is 27q^4 + 256r^3.

    Now, using the Symmetric Functions theorem, it's obvious that q= -s3 and r=s4 (where s3 and s4 are the third and fourth elementary symmetric polynomials). How do I proceed from here?

    Many thanks.
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  2. #2
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    Quote Originally Posted by KSM08 View Post
    Using the Theorem on Symmetric Functions or otherwise,
    show that the discriminant of the quartic polynomial x^4 + qx + r
    is 27q^4 + 256r^3.

    Now, using the Symmetric Functions theorem, it's obvious that q= -s3 and r=s4 (where s3 and s4 are the third and fourth elementary symmetric polynomials). How do I proceed from here?
    The discriminant is given by \Delta = \prod_{i<j}(x_i-x_j)^2, where x_1,\,x_2,\,x_3,\,x_4 are the roots of the polynomial (over the complex field, say).

    For a quartic polynomial, the discriminant is a symmetric function of the roots, of degree 12. But q has degree 3 and r has degree 4. The symmetric function theorem says that \Delta must be a function of the elementary symmetric functions, and in this example the only available such functions are q and r (because s_1 and s_2 are 0). So \Delta must be of the form mq^4+nr^3, for some constants m and n.

    To find m and n, evaluate \Delta in some special cases. For example, if q=0 and r=–1 then the equation becomes x^4=1, with roots \pm1,\,\pm i. From the definition of \Delta, you can check that \Delta = -256 in this case. So n=256. Similarly, if q = –1 and r=0 then the equation is x^4-x=0, with roots 0,\,1,\,(-1\pm\sqrt3i)/2. I haven't tried to check this, but presumably \Delta works out as 27 in this case.
    Last edited by Opalg; Jun 2nd 2010 at 02:11 PM. Reason: corrected error
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