# Solution of "polynomial" with integer and fractional powers

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1. Jun 15, 2016

### idmena

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. Jun 15, 2016

### stevendaryl

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

$x + a\sqrt{x^2 - b} + c = 0 \Rightarrow x+c = - a \sqrt{x^2 - b} \Rightarrow x^2 + 2cx + c^2 = ax^2 - ab$

3. Jun 15, 2016

### idmena

Of course, it's evident!

Thank you for your prompt answer.

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