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X in terms of a

  1. Jun 29, 2005 #1
    Hi,
    I have a problem with the following problem. We have to write x in terms of a.

    [tex]x^2=y+a[/tex]
    [tex]y^2=z+a[/tex]
    [tex]z^2=x+a[/tex]

    I have done some work but all I got is eighth degree polinomial and there is no easy wayof solving it in terms of a as far as I know. This is what I have

    [tex]((x^2-a^2)^2-a)^2=x[/tex]

    Any help is appreciated

    Cheers,
    Can
     
  2. jcsd
  3. Jun 29, 2005 #2

    OlderDan

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    That does not look quite right

    [tex] x^2 - a = y [/tex]
    [tex] y^2 - a = z [/tex]
    [tex] z^2 - a = x [/tex]

    [tex] \left( \left[ x^2 - a \right ]^2 - a \right )^2 - a = x \ \ \ \ ??? [/tex]
     
  4. Jun 29, 2005 #3
    Ummm, yes that's what I have but it seems I have written it wrong. I should have 8 roots x in terms of a in this case but I can only think of newton's method to find those. That is not a very good idea actually...
     
  5. Jun 29, 2005 #4
    Nice question,

    Think about using quadratic formula with x in terms of z and a.
     
  6. Jun 29, 2005 #5

    Zurtex

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    2 of the roots are simple quadratic roots but the other 6 can not be simplified any further than they are roots of an akward 6th degree polynomial.
     
  7. Jun 30, 2005 #6
    Do you mean something like this? English is not my native tongue so I have a little difficulty with mathematical terms. this is what I have now

    [tex]x^2+x-\sqrt (z+a) -z^2=0 [/tex]
     
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