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Thread: Polynomials

  1. #1
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    Polynomials

    Question:
    If $\displaystyle ax^3 + bx^2 + d=0$ has a double root, show that $\displaystyle 27a^2d+4b^3 = 0$

    I do not know how to start this question. However, I know that it uses this theorem:

    If a polynomial has a root $\displaystyle \alpha $ of multiplicity m, then P'(x) has the root $\displaystyle \alpha $ with multiplicity (m-1).

    Any help is appreciated.
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  2. #2
    Senior Member Peritus's Avatar
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    It's pretty straight forward:

    suppose r is the double root of the given polynomial, thus:

    $\displaystyle
    P(x = r) = ar^3 + br^2 + d = 0
    $

    now according to the theorem you've already stated:

    If a polynomial has a root \alpha of multiplicity m, then P'(x) has the root \alpha with multiplicity (m-1)
    $\displaystyle P'(x) = 3ax^2 + 2bx$

    so r must be one of the roots of P'(x), thus:

    $\displaystyle P'(r) = 3ar^2 + 2br = 0$


    $\displaystyle r\left( {3ar + 2b} \right) = 0$

    r=0 is a double root of P(x) only if d = 0, which is not necessarily true, so we'll choose: $\displaystyle r = - \frac{{2b}}{{3a}}$

    now substitute this value of r in the first equation:

    $\displaystyle
    a\left( { - \frac{{2b}}
    {{3a}}} \right)^3 + b\left( { - \frac{{2b}}
    {{3a}}} \right)^2 + d = 0
    $

    ... simplify and obtain the answer.
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  3. #3
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    Thank you very much . I was stuck on the r (3ar+2b) part. I never thought to find that root and substitute it.
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