roots of system of nonlinear equations

I can't get any solutions beside when $x=0\vee y=0 \vee z=0$ $$yz-2x\lambda-2x\mu=0\tag{1}$$ $$xz-2y\mu=0\tag{2}$$ $$xy-4z\lambda =0\tag{3}$$ $$x^2+y^2=4\tag{4}$$ $$x^2+2z^2=3\tag{5}$$ Can you help me? I seem to be doing algebraic manipulation that lead me nowhere.

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Using Mathematica wolframalpha.com/input/?i=Solve[{x^2+%2B+2*z^2+%3D%3D+3%2C+x^2+%2B+y^‌​2+%3D%3D+4%2C+xy+-+4*zl+%3D%3D+0%2C++++xz+-+2*yp+%3D%3D+0%2C+yz+-+2*xl+-+2*‌​x*p+%3D%3D+0}%2C+{x%2C+y%2C+z%2C+l%2C+p}], I get 16 different solutions for x,y,z,$\lambda$,$mu$. – NicoDean Jun 18 '14 at 12:48
thank you for your suggestion, but this system is what I have to solve for constrained optimization in exam, where I have to solve this kind of things without using electronic devices... – cgnx Jun 18 '14 at 12:50
Just saw that the Link isn't represented correctly. Copy the whole data until (without) the comma. – NicoDean Jun 18 '14 at 12:50

From (4) and (5) we obtain:

$$y^2z^2=(4-x^2)(3-x^2)/2......(6)$$

From (1) we obtain: $$y^2z^2=4x^2(\lambda+\mu)^2......(7)$$

From (6)-(7) we have: $$(4-x^2)(3-x^2)=8x^2(\lambda+\mu)^2...(8)$$

We can then solve $x^2$ from (8).

Once $x$ is solved, we can then solve other equations (4) and (5) for $y$ and $z$...

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I feel dumb, but how do I get x from (8)? – cgnx Jun 18 '14 at 13:31
(8) is a quadratic equation in $x^2$. – mike Jun 18 '14 at 13:34
I am sorry I still don't understand. If I had for example $(4-x^2)(3-x^2)=0$ I would know. But here to get 0 on one side I have to substract and I am left with $(4-x^2)(3-x^2)-8x^2(\lambda+\mu)^2=0$ – cgnx Jun 18 '14 at 13:38
So $x^2= (1/2)*(7 + 8*A^2 + \sqrt{1 + 112*A^2 + 64*A^4})$ where $A=(\lambda+\mu)^2$. – mike Jun 18 '14 at 13:43