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Solving a polynomial

  1. Apr 3, 2005 #1

    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.
  2. jcsd
  3. Apr 3, 2005 #2
    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: Apr 3, 2005
  4. Apr 3, 2005 #3

    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.
  5. Apr 3, 2005 #4


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    Well Mathematica can't simplify it.
  6. Apr 3, 2005 #5
    Agreed. That polynomial is in simplest form.
  7. Apr 3, 2005 #6
    it's not a polynomial, either.
  8. Apr 3, 2005 #7
    Thank you for giving us the correct definition of a polynomial.
    Last edited: Apr 3, 2005
  9. Apr 3, 2005 #8
    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.
  10. Apr 3, 2005 #9


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    Please don't spam. You already posted your homework question here.
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