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Math Help - how do you get roots of equations of fractional degree?

  1. #1
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    how do you get roots of equations of fractional degree?

    Say you have equations like 3x^(7/3) + 2x^(2/5) - 6 = 0 , the degree of the equation is 7/3..
    How do you solve for the roots? I know how to solve for the roots of higher degree equations like x^5, etc using rational root test but what about this? I ask because I encountered something like this in a calculus book. I made my calculator solve this and it gave me 1.1199, but I do not know if it is the only root or not.
    Last edited by catenary; April 23rd 2014 at 11:45 PM.
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  2. #2
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    Re: how do you get roots of equations of fractional degree?

    Quote Originally Posted by catenary View Post
    Say you have equations like 3x^(7/3) + 2x^(2/5) - 6 = 0 , the degree of the equation is 7/3..
    How do you solve for the roots? I know how to solve for the roots of higher degree equations like x^5, etc using rational root test but what about this? I ask because I encountered something like this in a calculus book. I made my calculator solve this and it gave me 1.1199, but I do not know if it is the only root or not.
    you can rewrite your expression as

    $3x^{7/3} + 2x^{2/5} - 6=0$

    $3x^{35/15} + 2x^{6/15}-6 =0$

    $u=x^{1/15}$

    $3u^{35} + 2u^6 - 6 = 0$

    This can be solved numerically by your root finding algorithm of choice. In this case there will be 35 solutions....

    The one real root is $u \approx 1.00758$

    That solution $u_0$ can then be used to find $x$

    $x=u^{15}$

    $x=(1.00758)^{15} \approx 1.11991$
    Thanks from catenary and topsquark
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    Re: how do you get roots of equations of fractional degree?

    Thanks romsek. Your explanation is as clear as could possibly be. This is one less situation I have to worry about not knowing how to deal with.
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