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X in terms of a

  • Thread starter wisredz
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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
 

OlderDan

Science Advisor
Homework Helper
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wisredz said:
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
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]
 
111
0
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...
 
Nice question,

Think about using quadratic formula with x in terms of z and a.
 

Zurtex

Science Advisor
Homework Helper
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wisredz said:
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...
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.
 
111
0
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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