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Math Help - Cubic Equations

  1. #1
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    Cubic Equations

    I have a cubic equation that I need to find a real solution to, does anyone have any ideas??

    The equation is ((A/B)-1)*(((A/C)^2)-1)=2*(A/C)*((D*E)/F)

    where -1 is actually -1 and not the inverse. All values are known except for that of A, this is the value that I need to find.

    Cheers,

    Ben
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  2. #2
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    Re: Cubic Equations

    rearange so you have F(a)=0, multiplying through by any denominators. If the values are known, can't you use those instead of B,C,D,E,F
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  3. #3
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    Re: Cubic Equations

    Well the issue is that I'm rearranging the formula to use in a VBA program, so I need to input the variables as B,C,D,etc as the values are calculated earlier in the program. So I need an equation which solves A in terms of B,C,D,etc. I'm just having an absolute nightmare in reformulating it. Keep coming to dead ends.

    (The value of F begins low and then increases until A is over a certain value)
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  4. #4
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    Re: Cubic Equations

    If those coefficients can be anything, then you will need the general "cubic" formula.

    Let a and b be any two numbers. (a+ b)^3= a^3+ 3a^2b+ 3ab^2+ b^3
    Also, 3ab(a+ b)= 3a^2b+ 3ab^3 so subtracting, we have (a+ b)^2- 3ab(a+ b)= a^3+ b^3

    If we let x= a+ b, m= 3ab, and n= a^3+ b^3, that equation is x^3- mx= n.

    Now, the question is, if we know m and n, can we solve for a and b- and so find x?

    From m= 3ab, we get b= m/3a. Putting that into a^3+ b^3= n, we have a^3+ \frac{m^3}{3^3a^3}= n. Multiply both sides by a^3 to get (a^3)^2+ \frac{m^3}{3^3}= na^3 or (a^3)^2- na^3+ \frac{m^3}{3^3}= 0 which we can think of as a quadratic equation for a^3.

    By the quadratic formula, a^3= \frac{n\pm\sqrt{n^2- 4\left(\frac{m}{3}\right)^3}}{2}.
    We can take that "2" in the denominator into the root and write
    a^3= \frac{n}{2}\pm\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^2}

    Because a^3+ b^3= n, b^3= n- a^3= \frac{n}{2}\mp\sqrt{\left(\frac{n}{2}\right)^2- \left(\frac{m}{3}\right)^2}

    That can be used to solve the "reduced cubic" with no " x^2" term.

    For the general cubic equation, ax^3+ bx^2+ cx+ d= 0, first let y= x+ q, for some unknown number q, so that
    ax^3+ bx^2+ cx+ d= a(y- q)^3+ b(y- q)^2+ c(y- q)+ d. Multiply that out and choose q so that the coefficient of y^2 is 0.
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