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Casus irreducibilis with positive roots

  1. Jun 17, 2011 #1
    Casus irredicibilis describes a situation where a cubic polynomial with integer coefficients has three distinct real roots, but you can't express those roots using just real numbers and radicals. Instead, if you want to only use radicals, then you must use complex numbers, even though the roots you're interested are purely real.

    The example that Planet Math gives is x^3 - 3x +1, but one of the roots of this cubic is negative. This is unsatisfying for me, because in the 1500's when Cardano first discovered the phenomena that cubic equations could not be solved with real numbers alone, the existence of negative numbers had not yet been commonly accepted. So in order to be more historically "authentic", I would prefer an example of casus irreducibilis where all three roots are positive. Does anyone know of such an example?

    Any help would be greatly appreciated.

    Thank You in Advance.
     
  2. jcsd
  3. Jun 17, 2011 #2
    A hackish solution: translate p(x) = x^3 - 3x + 1 over to the right by 2 to get p(x-2) = (x-2)^3 - 3(x-2) + 1 = x^3 - 6x^2 + 9x - 1. Now all your roots are positive :)
     
  4. Jun 17, 2011 #3
    How about an example of the form x^3 + ax + b?
     
  5. Jun 17, 2011 #4
    Say p(x) = x^3 + ax + b has 3 real roots. Then p''(x) = 6x, so the point of inflection of the cubic is at x = 0. There's no way 3 roots could occur to the right of the point of inflection. Also, looking at p'(x) = 3x^2 + a, the cubic has extrema located at x = ±√(-a/3). If p(-√(-a/3)) > 0 then there must be a (necessarily negative) root to the left of this maximum.

    In short, it's impossible.

    EDIT: Here is perhaps a more clear "proof" of the impossibility of your desired form: Let u, v, w in R be the three positive roots. Then p(x) = (x - u)(x - v)(x - w) = x^3 - (u + v + w) x^2 + (uv + uw + vw) x - uvw. It really just matters to consider the coefficient of x^2: if it is 0, then we have u + v + w = 0. All of u, v, w are positive, so this is only possible if u = v = w = 0.
     
    Last edited: Jun 17, 2011
  6. Jun 17, 2011 #5
    OK, thanks Unit, using similar reasoning we can easily see that none of the coefficients can be zero. But this leaves me puzzled: how would Cardano have come across a four-term cubic that had this special property?
     
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