Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Solve the system of equations: $$\begin{matrix} 2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}.y^2} \\ 2\sqrt[4]{\frac{y^4}{3}+4} = 1+\sqrt{\frac{3}{2}.x^2} \end{matrix}.$$

share|improve this question
    
Please update your equation. –  Patrick Li Jan 30 '13 at 6:32
6  
What have you tried? –  mixedmath Jan 30 '13 at 6:40
    
Upps, is this my homework-assignment for today? –  Gottfried Helms Jun 24 '13 at 9:25

2 Answers 2

The real solutions are $x = \pm \sqrt{6}$, $y = \pm \sqrt{6}$. There are also complex solutions.

EDIT: I found these using Maple's "solve" command. A somewhat more "hands-on" approach:

> with(Groebner):
> eqs:= [s^4-x^4/3 - 4,t^2-3/2*y^2,u^4-y^4/3-4,
       v^2-3/2*x^2,2*s-1-t,2*u-1-v];
> G:=Basis(eqs,plex(s,t,u,v,x,y));
> factor(G[1]);

$$\left( 1369\,{y}^{4}+660\,{y}^{2}+15876 \right) \\ ( 12065393290011975315218089\,{y}^{24}+31241483903922756527916216\,{y}^{ 22}\\+1916822411606153357786575368\,{y}^{20} + 2366873160958375355737537632\,{y}^{18}\\+108410617754815247626287499632 \,{y}^{16} +77702138286145060148486883072\,{y}^{14}\\+ 3076294432635247223638434223872\,{y}^{12} + 977270871111943893941425093632\,{y}^{10}\\+ 46998177944708847603452419729152\,{y}^{8} + 1724702245845492185473755174912\,{y}^{6}\\+ 369903763273618971503529112700928\,{y}^{4}- 28583510894171781210603713470464\,{y}^{2}\\+ 1174425086245411511666388854083584 ) \left( {y}^{2}-6 \right) ^ {2}$$

The first factor obviously has no real roots, the last has $y = \pm \sqrt{6}$. For the middle factor:

> sturm(sturmseq(op(2,%),y),y,-infinity,infinity);

$$0$$

So there are no real roots there either.

share|improve this answer
    
Thank you but you can able to complete? help me –  Maria Petrova Jan 30 '13 at 10:16

I will suppose we are only looking for real solutions.

Without loss of generality, we can assume that $x,y\ge0$. (We can than add signs to get the remaining solutions.)

So our equations are changed to

$$2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}}y\tag{1}$$ $$2\sqrt[4]{\frac{y^4}{3}+4}=1+\sqrt{\frac{3}{2}}x\tag{2}$$

If we subtract the two equations, we get (1)-(2): $$2\left(\sqrt[4]{\frac{x^4}{3}+4}-\sqrt[4]{\frac{y^4}{3}+4}\right)=y-x.\tag{3}$$ Notice that the function $x\mapsto\sqrt[4]{\frac{x^4}{3}+4}$ is increasing.

So for $x>y$ the LHS is positive and the RHS is negative. Similarly, for $y<x$ the signs of the two expressions are opposite.

So we can only find a real solution for $x=y$, which gives us $$2\sqrt[4]{\frac{x^4}{3}+4}=1+\sqrt{\frac{3}{2}}x\tag{4}$$

From this we get $$16\left(\frac{x^4}{3}+4\right)=\left(1+\sqrt{\frac{3}{2}}x\right)^4\tag{5}$$

This is a quartic equation. In theory, this can be done by hand, but it will be very probably quite messy. WolframAlpha returns this. (One of the solution, according to WA, is $\sqrt6$.)

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.