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Solution of "polynomial" with integer and fractional powers

  1. Jun 15, 2016 #1
    Hello, I have a question regarding "polynomials" that have terms with interger and fractional powers.
    1. The problem statement, all variables and given/known data
    I want to solve:
    $$ x+a(x^2-b)^{1/2}+c=0$$
    2. Relevant equations

    3. The attempt at a solution
    My approach is to make a change of variable x=f(y) to get a true polynomial (integer powers) that I know how to solve, e.g.:
    $$y^2+a y +b =0 $$
    Then I can switch back from y to x and use each of the solutions in y to get solutions in x. I find that works ok when the power of x inside the root is lower or equal than the power of x outside the root, for example, for the following equations:
    $$ x+a(x-b)^{1/2}+c=0 $$
    $$ x^2+a(x-b)^{1/2}+c=0 $$
    $$ x^2+a(x^2-b)^{1/2}+c=0 $$
    the solution is given by making the replacement: ##y=(x^n-b)^{1/2}##, then you have a polynomial in y, e.g.: ##y^2+ay+(b+c)=0## which is straightforward.

    If I try the same on this one I get a fractional power of y in the new polynomial, which puts me back in square 1. So far I have not been able to find the right change of variable for this problem.

    I am trying to work my way up to:
    $$ a x + b x^2 + c x^3 + d (e+fx+(g+hx+jx^2)^{1/2})^2 =0$$
    which is the actual equation I need to solve in the problem I am working on.

    Any hints?
    Thank you!
     
  2. jcsd
  3. Jun 15, 2016 #2

    stevendaryl

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    Staff Emeritus
    Science Advisor

    The way that I would do it is to try to get rid of the radical. For the simple example:

    [itex]x + a\sqrt{x^2 - b} + c = 0 \Rightarrow x+c = - a \sqrt{x^2 - b} \Rightarrow x^2 + 2cx + c^2 = ax^2 - ab[/itex]
     
  4. Jun 15, 2016 #3
    Of course, it's evident!

    Thank you for your prompt answer.
     
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