# Solving a polynomial

1. Apr 3, 2005

### jessawells

hi,

can anyone show me how to solve this:

$$\sqrt{x^3} + \sqrt{1+x^3}$$

i want to get it to so that there's only 1 term of x. but i don't know how to expand the squared root. any help is appreciated.

2. Apr 3, 2005

### Jameson

Usually when you say solve an equation, you mean something like solve x + 1 = 2.

Did you mean solve :$$\sqrt{x^3} = \sqrt{1+x^3}$$ (Probably not, since this is a false statement)

or simplify

$$\sqrt{x^3} + \sqrt{1+x^3}$$
?

I'm guessing the second one. Do you have an idea of how to approach it?

Last edited: Apr 3, 2005
3. Apr 3, 2005

### jessawells

hi,

yes, its the second case - simplifying the expression. i'm not sure at all how to approach it. i've thought about using the fact that both terms are squared - eg. making it $$\sqrt{x^3 + (1+x^3)}$$ but i know that's wrong. other than that, i've been trying to expand the $$\sqrt{1+x^3}$$ term, the way you would expand something like $$(1+x)^2$$, but i've had no success. any help would be great.

4. Apr 3, 2005

### Zurtex

Well Mathematica can't simplify it.

5. Apr 3, 2005

### Jameson

Agreed. That polynomial is in simplest form.

6. Apr 3, 2005

### Data

it's not a polynomial, either.

7. Apr 3, 2005

### Jameson

Thank you for giving us the correct definition of a polynomial.

Last edited: Apr 3, 2005
8. Apr 3, 2005

### Data

No... polynomials only have natural powers. A polynomial is something of the form

$$\alpha_0 + \alpha_1x + \alpha_2x^2 + \ . \ . \ . \ + \alpha_nx^n.$$

which $\sqrt{x^3} + \sqrt{x^3+1}$ certainly is not.

Now, you can certainly do some interesting things to $\sqrt{x^3} + \sqrt{x^3 + 1}$, as usual. For example,

$$\sqrt{x^3} + \sqrt{x^3 + 1} = \frac{1}{\sqrt{x^3+1}-\sqrt{x^3}}$$

$$= \sqrt{2\sqrt{x^3}\left(\sqrt{x^3}+\sqrt{1+x^3}\right) + 1}$$

but I would by no means consider those simpler.

9. Apr 3, 2005

### Hurkyl

Staff Emeritus