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Roots of cubic equation

  1. Aug 12, 2011 #1
    1. The problem statement, all variables and given/known data

    θ^3 - pθ^2 +qθ - r = 0 such that p and r do not equal zero

    If the roots can be written in the form ak^-1, a, and ak for some constants a and k, show that one root is q/p and that q^3 - rp^3 = 0. Also, show that if r=q^3/p^3, show that q/p is a root and that the product of the other roots is (q/p)^2

    2. Relevant equations



    3. The attempt at a solution

    Mind boggling, can anyone give me so much as a hint?
     
  2. jcsd
  3. Aug 12, 2011 #2

    SammyS

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    What do you mean by ak^-1 ?

    Do you mean a/k , OR do you mean 1/(ak) ?
     
  4. Aug 12, 2011 #3

    SammyS

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    If the roots are x , y, and z , then

    (θ -x)(θ - y)(θ - z) = θ3 - pθ2 +qθ - r
     
  5. Aug 12, 2011 #4
    I am hazarding a guess that he means a/k, so that the roots form a geometric progression (which was usually the basis for a lot of 'roots of polynomial' questions I used to get)
     
  6. Aug 12, 2011 #5
    With a cubic expression in the form:
    \begin{align}
    ax^3 + bx^2 + cx + d \\
    \end{align}
    If we call the roots [itex]\alpha[/itex], [itex]\beta[/itex] and [itex]\gamma[/itex] then:
    \begin{align}
    a(x-\alpha)(x-\beta)(x-\gamma) = ax^3 + bx^2 + cx + d \\
    \end{align}

    Where:
    \begin{align}
    \alpha + \beta + \gamma = \frac{-b}{a} \\

    \alpha\beta + \beta\gamma + \gamma\alpha= \frac{c}{a} \\

    \alpha\beta\gamma= \frac{-d}{a} \\
    \end{align}

    In this specific case, a = 1. That lot should help you!
    Also please notice that the 'a' I use to represent the coefficient of x cubed is not the same as the 'a' you have been given in the question.


    Extra Hint (Only use if you are still stuck):




    \begin{align}
    \alpha\beta\gamma = \frac{a}{k} \times a \times ak = a^3
    \end{align}
     
    Last edited: Aug 12, 2011
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