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I need to solve this nonlinear system of equations. Coefficients a,b,c,d are given. Since it is a system of 4 eq. and 4 unknowns $x_{i}$ it should have a solution but unfortunately there are the squares... Is there any way to solve this in relatively elegant way? I tried but only get a mess...

$ax_{1}^2 + 2bx_{1}x_{3} + cx_{3}^2-a=0$

$ax_{1}x_{2} + bx_{1}x_{4} + bx_{2}x_{3} + cx_{3}x_{4} - b=0$

$ax_{2}^2+2bx_{2}x_{4} +cx_{4}^2 -c= 0$

$x_{1}x_{4}-x_{2}x_{3}=1$

Mark
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  • Consider replacing X1 by the last equation ? – mick Jan 04 '18 at 22:15
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    @mick how can you replace an unknown by an equation ? You mean using the last equation to express $x_1=\dfrac{1+x_2x_3}{x_4}$ as a function of the other unknowns? – Jean Marie Jan 04 '18 at 22:18
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    Just a quick thought: your first and third equations are quadratic and can be solved for one of the variables. Substituting the result of that into the fourth might be useful. You should mention what you've tried already in your question to help answerers. – Sort of Damocles Jan 04 '18 at 22:20
  • Yes and together with dbx comment that might work – mick Jan 04 '18 at 22:28
  • One apparent solution is $\vec x = \pm(1,0 , 0, 1)$ – lulu Jan 04 '18 at 22:38
  • Hee again...I obtain a discriminant with one variable and three coefficients under the sqrt that must be inserted in the remaining eqs. Replacing $x_{1}$ and $x_{2}$ from the first and third eq. in the fourth I get $+- x_{4} \sqrt {x_{3}^2 (b^2-ac)+a^2} -+x_{3} \sqrt {x_{4}^2 (b^2-ac)+ac}=a $. Now one should solve for $x_{3}$ or $x_{4}$ and insert in the second eq. Seems to me messy... or what am I doing wrong? – Mark Jan 04 '18 at 23:29
  • there isn't a general method of doing this. If you don't need to do it by hand, look for a copy of mathematica or similar. – Sort of Damocles Jan 05 '18 at 04:17
  • Mathematica explodes it into a horribly long expression (but this does not necessarily mean that there isn't a simpler format). Maybe there is no other way than to do it numerically or simply forget about it. I just wanted to be sure that there is no way to do that and that it isn't me who couldn't see the proper method. – Mark Jan 05 '18 at 12:55

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