Solving a polynomial

  • Thread starter jessawells
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  • #1
19
0
hi,

can anyone show me how to solve this:

[tex]\sqrt{x^3} + \sqrt{1+x^3}[/tex]

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.
 

Answers and Replies

  • #2
789
4
Usually when you say solve an equation, you mean something like solve x + 1 = 2.

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

or simplify

[tex]\sqrt{x^3} + \sqrt{1+x^3}[/tex]
?

I'm guessing the second one. Do you have an idea of how to approach it?
 
Last edited:
  • #3
19
0
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 [tex]\sqrt{x^3 + (1+x^3)}[/tex] but i know that's wrong. other than that, i've been trying to expand the [tex]\sqrt{1+x^3}[/tex] term, the way you would expand something like [tex](1+x)^2[/tex], but i've had no success. any help would be great.
 
  • #4
Zurtex
Science Advisor
Homework Helper
1,120
1
Well Mathematica can't simplify it.
 
  • #5
789
4
Agreed. That polynomial is in simplest form.
 
  • #6
998
0
it's not a polynomial, either.
 
  • #7
789
4
Thank you for giving us the correct definition of a polynomial.
 
Last edited:
  • #8
998
0
No... polynomials only have natural powers. A polynomial is something of the form

[tex]\alpha_0 + \alpha_1x + \alpha_2x^2 + \ . \ . \ . \ + \alpha_nx^n.[/tex]

which [itex]\sqrt{x^3} + \sqrt{x^3+1}[/itex] certainly is not.

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

[tex] \sqrt{x^3} + \sqrt{x^3 + 1} = \frac{1}{\sqrt{x^3+1}-\sqrt{x^3}}[/tex]

[tex] = \sqrt{2\sqrt{x^3}\left(\sqrt{x^3}+\sqrt{1+x^3}\right) + 1}[/tex]

but I would by no means consider those simpler.
 
  • #9
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,916
19
Please don't spam. You already posted your homework question here.
 

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