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Polynomial Equation

  1. Jan 5, 2008 #1
    I'd like to know how to find relations between coefficients of a polynomial p(x) so that equation p(x) = 0 has only real roots.

    For example I have quadratic equation:

    x[tex]^{2}[/tex]+px+q = 0

    then the Discriminant must be >= 0 so p[tex]^{2}[/tex] >= 4q

    But I need to find it for equations of higher power of x (like 6), without having the formula for exact solutions - I know it's not possible to solve polynomial equations of order 6 and more.

    It may be complicated if there are too many coefficients (there are 5 in 6th order pol.), but they just all depend just on one constant r, so it is like a5 = 2r, a4 = r[tex]^{2}[/tex], a3 = 1/r, and so on, and I need to know for which values of r the equation has only real roots.
  2. jcsd
  3. Jan 5, 2008 #2


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    First you had better clear up some misconceptions. In the first place, a 6[/sup]th[/sup] degree (not "order") equation would have 7 coefficients (or 6 if you always take the leading coefficient to be 1 and ignore it), not 5. In general, an nth degree equation has n+1 coefficients, not n-1. Again, you could always divide the equation by the leading coefficient and just use the other n.
    Secondly, any set of number can be written as functions of one parameter. How many real roots an equation has would depend upon how each coefficient depends on r and that might be very complicated. Even in the very simple case [itex]x^3+ bx^2+ cx+ d[/itex] whether there are non-real roots would depend upon whether the local min and max are on opposite sides of the x-axis- and that alone can be complicated.
  4. Jan 6, 2008 #3
    i dont know about all real roots ,but i can tell you that for all complex roots
    in equation of n th degree whose first coefficient is 1
    second ceofficient sqaure<2*third coefficient
    and if it is = then all the roots are 0
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