I am trying to prove that the polynomial f=X3+pX+q∈R[X] has three real roots iff Δ(f)=−(4p3+27q2)≥0. I've figured out the left-to-right direction. Write f=(X−α1)(X−α2)(X−α3), and then notice that
Δ(f)=(α1−α2)2(α1−α3)2(α2−α3)2≥0.
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(α1−α2)2(α1−α3)2(α2−α3)2=(α1−α2)2(α1−¯α2)2(α2−¯α2)2=|α1−α2|4(2iIm(α2))2=−4|α1−α2|4(Im(α2))2<0.
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