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  • OT Maths transformation.

    OT Maths transformation.

    Hi I want to transpose the following Formula to give X.

    30 years ago at night school I could do this sort of thing now days I’m stuck, can any one help and show a step by step transformation.

    Y=a+b*x+c/x^2

    I think first step is
    y-a = b*x+c/x^2

    next step I go blank

    Steve

  • #2
    You can get this to a standard cubic equation by multiplying by x^2 and rearranging:

    x^3-((y-a)/b)x^2+c/b=0

    Then there is a formula to find roots of cubics, but you're not going to like it:

    http://www.math.vanderbilt.edu/~schectex/courses/cubic/
    For just a little more, you can do it yourself!

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    • #3
      Marv--

      You may have missed the "divide" slash in the last term. This makes the exponents in the equation span from -2 to 1. Multiplying through by x^2 makes the span 0 to 3.
      For just a little more, you can do it yourself!

      Comment


      • #4
        You're right. I missed that. I've deleted the errant post. Sorry about that.
        Regards, Marv

        Home Shop Freeware - Tools for People Who Build Things
        http://www.myvirtualnetwork.com/mklotz

        Location: LA, CA, USA

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        • #5
          Well... you sure stumped me. I can get the answer easy enough by plugging into Wolfram Alfa, but the steps to get there are unknown to me. When the button for showing "intermediate steps" is clicked on Wolfram Alfa, it is conveniently "unavailable". One answer shows an i-root. If I remember correctly from the only math class I ever failed, that "i" is calculus speak for "imaginary numbers." Uh-oh.
          Last edited by Arthur.Marks; 07-26-2012, 03:18 PM.

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          • #6
            The * after the a+b* signifies what, unknown power or multiplication? Peter
            The difficult done right away. the impossible takes a little time.

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            • #7
              The * signifies multiplication. It's the same as stating a+bx.
              I entered the formula in Microsoft Mathematics and it solved for X automatically but I can't seem to copy and paste here. I'm too lazy to create an image. Download the free version of MM and run it. It will even plot it for you in 2D or 3D.

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              • #8
                see
                http://en.wikipedia.org/wiki/Quadratic_equation

                Comment


                • #9
                  I get:
                  [x=(-(sqrt(3)*%i)/2-1/2)*(sqrt(-c*(4*y^3-12*a*y^2+12*a^2*y-27*b^2*c-4*a^3))/(2*3^(3/2)*b^2)-(-2*y^3+6*a*y^2-6*a^2*y+27*b^2*c+2*a^3)/(54*b^3))^(1/3)+(((sqrt(3)*%i)/2-1/2)*(y^2-2*a*y+a^2))/(9*b^2*(sqrt(-c*(4*y^3-12*a*y^2+12*a^2*y-27*b^2*c-4*a^3))/(2*3^(3/2)*b^2)-(-2*y^3+6*a*y^2-6*a^2*y+27*b^2*c+2*a^3)/(54*b^3))^(1/3))-(a-y)/(3*b),

                  x=((sqrt(3)*%i)/2-1/2)*(sqrt(-c*(4*y^3-12*a*y^2+12*a^2*y-27*b^2*c-4*a^3))/(2*3^(3/2)*b^2)-(-2*y^3+6*a*y^2-6*a^2*y+27*b^2*c+2*a^3)/(54*b^3))^(1/3)+((-(sqrt(3)*%i)/2-1/2)*(y^2-2*a*y+a^2))/(9*b^2*(sqrt(-c*(4*y^3-12*a*y^2+12*a^2*y-27*b^2*c-4*a^3))/(2*3^(3/2)*b^2)-(-2*y^3+6*a*y^2-6*a^2*y+27*b^2*c+2*a^3)/(54*b^3))^(1/3))-(a-y)/(3*b),

                  x=(sqrt(-c*(4*y^3-12*a*y^2+12*a^2*y-27*b^2*c-4*a^3))/(2*3^(3/2)*b^2)-(-2*y^3+6*a*y^2-6*a^2*y+27*b^2*c+2*a^3)/(54*b^3))^(1/3)+(y^2-2*a*y+a^2)/(9*b^2*(sqrt(-c*(4*y^3-12*a*y^2+12*a^2*y-27*b^2*c-4*a^3))/(2*3^(3/2)*b^2)-(-2*y^3+6*a*y^2-6*a^2*y+27*b^2*c+2*a^3)/(54*b^3))^(1/3))-(a-y)/(3*b)]

                  As the three roots (courtesy of Maxima.)

                  bob_s,

                  Though one can use the Quadratic equation to solve this it is not a quadratic. It is cubic. The formulas and strategies for solving cubic functions are here:
                  http://en.wikipedia.org/wiki/Cubic_function

                  I can manipulate it to look like this:
                  X^2 * (bx - a - y) + c = 0

                  -DU-

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                  • #10
                    Originally posted by SDL View Post
                    Hi I want to transpose the following Formula to give X.

                    y=a+b*x+c/x^2
                    This is an equation of y as a function of x, or y=f(x). For this particular form there is no explicit function g such that x=g(y).

                    But you can find the roots, which are the values for which y=0. You then have the cubic ax^2+bx^3+c = 0, which can be solved as described here: http://en.wikipedia.org/wiki/Cubic_equation
                    Last edited by aostling; 07-26-2012, 09:18 PM. Reason: mis-multiplied and got the wrong equation
                    Allan Ostling

                    Phoenix, Arizona

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