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Math Help - Cubic equation

  1. #1
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    Cubic equation



    I'm not even sure where to begin... Help will be appreciated
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  2. #2
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    Re: Cubic equation

    $(x-a)^2(x-b) = x^3+px+q$

    Multiply out:

    $(x-a)^2(x-b) = (x^2-2ax+a^2)(x-b) = x^3-(2a+b)x^2+(2ab+a^2)x-a^2b$

    Equate coefficients:

    $-(2a+b) = 0$

    $2ab+a^2=p$

    $-a^2b = q$

    This is a system of three equations in four variables. So, solve for $p$ in terms of $q$.
    Last edited by SlipEternal; April 14th 2014 at 06:39 AM.
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  3. #3
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    Re: Cubic equation

    $x^3 + px + q = (x - a)^2(x - b) = (x^2 - 2ax + a^2)(x - b) = x^3 - bx^2 - 2ax^2 + 2abx + a^2x - a^2b =$

    $x^3 - (2a + b)x^2 + a(a + 2b)x - a^2b.$ Straight forward.

    The coefficients of the powers must be equal in the right hand and left expressions.

    $\therefore q = - a^2b\ and\ p = a(a + 2b)\ and\ 2a + b = 0.$

    Proceed
    Last edited by JeffM; April 14th 2014 at 06:21 AM. Reason: LaTeX gave me fits
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  4. #4
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    Re: Cubic equation

    Quote Originally Posted by JeffM View Post
    $x^3 + px + q = (x - a)^2(x - b) = (x^2 - 2ax + a^2)(x - b) = x^3 - bx^2 - 2ax^2 + 2abx + a^2x - a^2b =$

    $x^3 - (2a + b)x^2 + a(a + 2b)x - a^2b.$ Straight forward.

    The coefficients of the powers must be equal in the right hand and left expressions.

    $\therefore q = - a^2b\ and\ p = a(a + 2b)\ and\ 2a + b = 0.$

    Proceed
    I'm being really stupid and going wrong somewhere.

    The answer is:

    but I'm not getting that...
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  5. #5
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    Re: Cubic equation

    Quote Originally Posted by JeffM View Post
    <snip>
    $\therefore q = - a^2b\ and\ p = a(a + 2b)\ and\ 2a + b = 0.$
    Quote Originally Posted by Bude8 View Post
    The answer is:
    JeffM and SlipEternal have given you 3 equations that must be satisfied

    $q = - a^2b\ and\ p = a(a + 2b)\ and\ 2a + b = 0.$

    from the last one $b=-2a$

    substitute that into the first two and you get

    $q=2a^3, ~p=-3a^2$

    $q^2=4a^6$

    $\dfrac{q^2}{4}=a^6$

    $p^3=-27a^6$

    $-\dfrac{p^3}{27}=a^6$

    $\dfrac{q^2}{4}=-\dfrac{p^3}{27}$

    $\dfrac{q^2}{4}+\dfrac{p^3}{27}=0$

    $27q^2+4p^3=0$
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