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Irrational polynomial equation

  1. Oct 19, 2007 #1
    what's the polynomial equation which sqrt2 + sqrt3 satisfies ?
  2. jcsd
  3. Oct 19, 2007 #2
    How about x-sqrt2-sqrt3?
  4. Oct 20, 2007 #3
    There is not one unique polynomial that (sprt2 + sqrt3) satisfies. Perhaps you should elaborate a bit as to what terms, degree, etc. that you require in your polynomial.
  5. Oct 20, 2007 #4


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    And keep in mind that a polynomial is nothing more than a linear combination of monomials...
  6. Oct 21, 2007 #5

    Gib Z

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    What about, [tex]x- \pi[/tex] ? or [tex]x^3 - 31[/tex] =] ?
  7. Oct 21, 2007 #6
    If you want to get rid of the roots, then it'd be something like

    [tex]x = \sqrt{2}+\sqrt{3}[/tex]
    [tex]x^{2} = 2+3+2\sqrt{6}[/tex]
    [tex]x^{2}-5 = 2\sqrt{6}[/tex]
    [tex](x^{2}-5)^{2} = 24[/tex]

    Rearrange and clean up a bit.
  8. Oct 21, 2007 #7
    yeah, i seems alright that way. well, i has to be polynomial, but i couldn't just go upto that part, so i was confused. thnx anyway :)
  9. Oct 24, 2007 #8


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  10. Oct 24, 2007 #9
    Hurkyl: And keep in mind that a polynomial is nothing more than a linear combination of monomials...

    From the standpoint of Galois Theory, we can build that form from the conjugates of the two order two equations, which produces an order 4 equation:

    The four products of the form: [tex]\prod (X-(\pm\sqrt2\pm\sqrt3)[/tex] = X^4-10X^2+1.
    Last edited: Oct 24, 2007
  11. Oct 25, 2007 #10


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    It would have been nice if you had posted the actual question. I suspect it was NOT "Find a polynomial equation that [itex]\sqrt{2}+ \sqrt{3}[/itex] satisfies". I suspect rather that it was something like "Find a polynomial equation, with integer coefficients, that [itex]\sqrt{2}+ \sqrt{3}[/itex] satisfies".
  12. Oct 25, 2007 #11

    Gib Z

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    I'm just joking with you man, try it on yours calculator, [itex]\sqrt{2}+\sqrt{3}[/itex] is a valid approximation of pi to several digits. Similar thing for the other equation.
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