Friday, March 3, 2017

galois theory - Polynomial has real roots



I am trying to prove that the polynomial f=X3+pX+qR[X] has three real roots iff Δ(f)=(4p3+27q2)0. I've figured out the left-to-right direction. Write f=(Xα1)(Xα2)(Xα3), and then notice that
Δ(f)=(α1α2)2(α1α3)2(α2α3)20.
But does anyone have a hint for the other direction? Suppose the discriminant is 0, how to prove f has three real roots (not necessarily distinct)?


Answer



What's the alternative? Well, your polynomial can have one real root α1 and two non-real conjugate roots α2 and α3=¯α2. Therefore(α1α2)2(α1α3)2(α2α3)2=(α1α2)2(α1¯α2)2(α2¯α2)2=|α1α2|4(2iIm(α2))2=4|α1α2|4(Im(α2))2<0.



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