Equation: (x^3 + k*x)^(1/5) = (x^5 – k*x)^(1/3)

 

Find a rational value of k such that there are four nonzero rational roots to the equation

 
EQUATION RATIONAL K

 
 
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
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2 Responses to Equation: (x^3 + k*x)^(1/5) = (x^5 – k*x)^(1/3)

  1. pipo says:

    For k =12: (x^3+12x)^(1/5) = (x^5-12x)^(1/3) has a solution for x =2.
    For k =72: (x^3+72x)^(1/5 )= (x^5-72x)^(1/3) has a solution for x =3.
    For k = 240: (x^3+240x)^(1/5 )= (x^5-240x)^(1/3) has a solution for x =4.
    For k = 600: (x^3+600x)^(1/5) = (x^5-600x)^(1/3) has a solution for x =5.

    So for every k = (x-1)*(x+1)*x^2 it has a unique solution.
    pipo

  2. pipo says:

    Oops, misread. I see I had to find four nonzero roots for a certain k. More complicated then I thought.
    pipo

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