# How do you isolate a variable in a function such as this

x^4+y^4+z^4+(4x^2)*(y^2)*(z^2)-34=0
ive tried all sorts of things factoring, algebraic manipulation ect. but i dont get how you could isolate a variable for a function like this i.e z= f(x,y).

You may not be able express one variable explicitly in terms of the other two.

Everything seems to be raised to powers of two. Therefore for some variable say x, replace x^2 with w, and then find w using the quadratic equation.

HallsofIvy
Homework Helper
x^4+y^4+z^4+(4x^2)*(y^2)*(z^2)-34=0
ive tried all sorts of things factoring, algebraic manipulation ect. but i dont get how you could isolate a variable for a function like this i.e z= f(x,y).
If you want to solve for z (isolate z), treat it as a polynomial is z. For example, write it as $z^4+ [4x^2y^2]z^2+ [x^4+ y^4- 34]= 0$
That's a fourth degree equation but, as John Chrieto suggests, since it has only even powers of z, letting $w= z^2$ gives $w^2+ [4x^2y^2]w+ [x^4+ y^4- 34]= 0$, a quadratic equation that can be solved using the quadratic formula- giving two solutions in terms of x and y. Then you can take the two square roots of each of those to give the four solutions to the fourth degree equation.

wow thank you guys you went well and beyond what i was thinking i didnt even think to substitute any of the squared variables and solve as a qudratic. Great insight/ intellegence.

ok so i tried doing that but i get stuck at this point: i did w^2+w*u+v=0 where w=z^2 and u and v are both equal to the separated "clump" functions. then i plug the variables into the quadratic equation and substitute and i get radical (x^4*(4y^4-1)-y^2-34)-2x^2*y^2 i don't get how you get two functions one in x and one in y.

HallsofIvy