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Math Help - Solving difficult equation

  1. #1
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    Solving difficult equation

    Hello everyone

    I have the following equation:

    y(x)=a_{1}x+a_{2}x^{C}

    where a_{1},a_{2} are constants, and C>0 is a real number (normally is not an integer). I would like to know ALL possible methods
    (mainly analytical) for solving this equation.

    Thanks
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  2. #2
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    Quote Originally Posted by JulieK View Post
    Hello everyone

    I have the following equation:

    y(x)=a_{1}x+a_{2}x^{C}

    where a_{1},a_{2} are constants, and C>0 is a real number (normally is not an integer). I would like to know ALL possible methods
    (mainly analytical) for solving this equation.

    Thanks
    Are you assuming that y = 0?
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  3. #3
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    Quote Originally Posted by JulieK View Post
    Hello everyone

    I have the following equation:

    y(x)=a_{1}x+a_{2}x^{C}

    where a_{1},a_{2} are constants, and C>0 is a real number (normally is not an integer). I would like to know ALL possible methods
    (mainly analytical) for solving this equation.

    Thanks

    There's no equation here at all. There's a function , y(x) , so there's nothing to solve here.

    A equation could be, for example, a_1x+a_2x^C=0 ...

    Tonio
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  4. #4
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    No I am not assuming this in general. However, what would you suggest if it is zero.
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  5. #5
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    Quote Originally Posted by JulieK View Post
    No I am not assuming this in general. However, what would you suggest if it is zero.
    Factorise, then use the Null Factor Law.


    In any other case, where y is a number, move everything to one side, then use the Factor and Remainder Theorems (if possible).
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  6. #6
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    Quote Originally Posted by tonio View Post
    There's no equation here at all. There's a function , y(x) , so there's nothing to solve here.

    A equation could be, for example, a_1x+a_2x^C=0 ...

    Tonio
    Let's assume this is a polynomial but with C non-integer. Then what I want is to find all x values that solve this equation as in the case of a polynomial. You can call it function, or equation or expression or anything else. This is not my point.
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  7. #7
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    Quote Originally Posted by Prove It View Post
    Are you assuming that y = 0?
    No I am not assuming this in general. However, what would you suggest if it is zero.
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  8. #8
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    Quote Originally Posted by Prove It View Post
    Factorise, then use the Null Factor Law.


    In any other case, where y is a number, move everything to one side, then use the Factor and Remainder Theorems (if possible).
    But y is not zero in genral (in fact in most cases y>0).
    Also I don't think Factor and Remainder Theorems is relevant.
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  9. #9
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    Quote Originally Posted by JulieK View Post
    But y is not zero in genral (in fact in most cases y>0).
    Also I don't think Factor and Remainder Theorems is relevant.
    In order to solve any nonlinear equation, usually the only way is to factorise and use the Null Factor Law. The easiest way to factorise is through use of the Factor and Remainder Theorems.
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  10. #10
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    I re-state my problem:

    Solve the following equation for x:

    a_{1}x^{\beta}+a_{2}x+a_{3}=0

    where a_{1},a_{2},a_{3} are constants, and \beta>0 is a real number which, for practical reasons, can be assumed to be rational.
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  11. #11
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    Quote Originally Posted by JulieK View Post
    I re-state my problem:

    Solve the following equation for x:

    a_{1}x^{\beta}+a_{2}x+a_{3}=0

    where a_{1},a_{2},a_{3} are constants, and \beta>0 is a real number which, for practical reasons, can be assumed to be rational.

    I don't think the above equation can, in general, be solved in terms of elementary functions, not even if \beta\in\mathbb{N} , and much worse if it isn't a natural number.

    In general, and with some additional assumptions, perhaps you can prove there's (or there isn't) a solution (or solutions), but the actual ones I don't think so.

    Tonio
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