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Need proof that a cubic equation has at least one real root

  1. Apr 22, 2005 #1

    A cubic equation has at least one real root.
    Can someone help me to prove this?

  2. jcsd
  3. Apr 22, 2005 #2


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    U need calculus.It's not difficult.Use the continuitiy on [itex] \mathbb{R} [/itex] and evaluate the 2 asymptotic limits.Then use the Rolle (IIRC) theorem...

  4. Apr 22, 2005 #3

    matt grime

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    I don't think you necessarily need that.

    If w is a complex root of f(x) a real polynomial (ie one with real coeffs), then so is w* the conjugate of w, this means that the real poly (x-w)(x-w*) divides f(x) over the reals. Hence complex roots occur in conjugate pairs. A cubic has 3 (possibly complex) roots, so pairing up the complex ones (or any poly of odd degree) means there must be an odd number of unpaired real roots left.
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