# X in terms of a

#### wisredz

Hi,
I have a problem with the following problem. We have to write x in terms of a.

$$x^2=y+a$$
$$y^2=z+a$$
$$z^2=x+a$$

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

$$((x^2-a^2)^2-a)^2=x$$

Any help is appreciated

Cheers,
Can

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#### OlderDan

Homework Helper
wisredz said:
Hi,
I have a problem with the following problem. We have to write x in terms of a.

$$x^2=y+a$$
$$y^2=z+a$$
$$z^2=x+a$$

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

$$((x^2-a^2)^2-a)^2=x$$

Any help is appreciated

Cheers,
Can
That does not look quite right

$$x^2 - a = y$$
$$y^2 - a = z$$
$$z^2 - a = x$$

$$\left( \left[ x^2 - a \right ]^2 - a \right )^2 - a = x \ \ \ \ ???$$

#### wisredz

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...

#### NewScientist

Nice question,

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

#### Zurtex

Homework Helper
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.

#### wisredz

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

$$x^2+x-\sqrt (z+a) -z^2=0$$

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