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Math Help - My big problem

  1. #1
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    My big problem

    Hello
    I have a problem with a function,
    Is it posible to calculate it, if hte exp is 10 or
    maybe a number between 6 and 10 (6,7,8,9,10)

    y = P - A x + B x^2 C y^a

    How can I find y?

    P,A,B,C,a are constants

    Thank you very much
    Last edited by mr fantastic; July 20th 2009 at 02:11 AM. Reason: Restored original question.
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  2. #2
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    Quote Originally Posted by Rinol View Post
    Hello
    I have a problem with a function,


    y = P - A x + B x^2 C y^a

    How can I find y?

    P,A,B,C,a are constants

    Thank you very much
    If a= 2 that is a quadratic equation in y and you can use the quadratic formula. If a= 3 or 4, it is a cubic or quartic equation, respectively, and there are formulas for solving for y (but much more complicated than the quadratic equation!). If a= 1/2, you can let z= y^(1/2) then you get z^2= P- Ax+ Bx^2 z, a quadratic equation in z. You can use the quadratic formula to solve for z and then take the square root find y. If a= 1/3 or 1/4, similarly, let z= y^(1/3) or z= y^(1/4) respectively to get polynomial equations that can be solved for z.

    If a not one of those numbers, there are no general formulas for solving such an equation.
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  3. #3
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    Quote Originally Posted by Rinol View Post
    y = P - A x + B x^2 C y^a
    Are you sure you posted that properly?

    Could the "y" on the left be a typo?
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  4. #4
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    My big problem

    Hallo everybody,

    I posted my problem last week
    but now I know my exponent, it is 10.

    I need to get the therm for y!

    y = P - Ax + BC x^2 y^10

    P,A,B,C are constants!

    Thanks a lot
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  5. #5
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    Quote Originally Posted by Rinol View Post
    I need to get the therm for y!
    y = P - Ax + BC x^2 y^10
    Let u = BC x^2 and v = P - Ax; then:
    y = v + u y^10
    u y^10 - y = -v
    y(u y^9 - 1) = -v

    I see iteration as the only way to solve this
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