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Solution of "polynomial" with integer and fractional powers

  • #1
14
0
Hello, I have a question regarding "polynomials" that have terms with interger and fractional powers.

Homework Statement


I want to solve:
$$ x+a(x^2-b)^{1/2}+c=0$$

Homework Equations



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!
 

Answers and Replies

  • #2
stevendaryl
Staff Emeritus
Science Advisor
Insights Author
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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]
 
  • #3
14
0
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]
Of course, it's evident!

Thank you for your prompt answer.
 

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