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Thread: Prove general solution for deppresed cubic equation y^3+py+q=0 given a discriminant

  1. #1
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    Smile Prove general solution for deppresed cubic equation y^3+py+q=0 given a discriminant

    I've been given the following problem and I can't seem to figure out the solution, would appreciate any help and direction to solution !

    Let's look at the equation : y3 + py + q = 0 (*)
    We shall define Delta (or d in short) d = 4p3 + 27q2
    Prove that :
    a) If d > 0 , then (*) has a single solution .

    b) If d = 0 , and at least one of the coefficients (p , q) =/= 0, then (*) has 2 solutions .

    c) If d < 0 , then (*) has 3 solutions .



    Thank you very much in advance !

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  2. #2
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    Re: Prove general solution for deppresed cubic equation y^3+py+q=0 given a discrimina

    if $r$ is a real root then

    $y^3+p y + q=(y-r)\left(y^2+r y+r^2+p\right)$

    the existence of additional real roots depends on the discriminant of the quadratic factor

    use the identity

    $\left(3r^2+4p\right)\left(p+3r^2\right)^2=4 p^3+27 q^2$
    Thanks from topsquark
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