I am developing a solution for a Chemical Engineering course, and I came across an interesting (tricky) equation system, for which I am curious if it is possible to solve it explicitly (closed form solutions). I am not interested in the solution "per-se" as I could just use MATLAB or any other program to get such solutions.

I have the following system of 2 nonlinear equations in the 2 unknowns $\xi$ and $\eta$: $$ \begin{cases} A(1-\xi)(B+\xi)=(C+\eta)(D+\eta)\\[3pt] E(1-\xi)(B+\xi)=(F+\xi-\eta)(G+\xi-\eta) \end{cases} $$ where $A,B,C,D,E,F,G$ are all strictly positive reals. The conditions of existence of $\xi$ and $\eta$ are: $$ 0<\eta<\xi\le1 $$ Is it possible to rearrange/transform, or more generally exploit the form of the two equations to obtain $\xi$ and $\eta$ in a closed form? I have tried to do so, but to no avail.

The only thing I noticed is that the first equation is in the form $f(\xi)=g(\eta)$, therefore the only condition is that both functions are equal to a constant. However, I have not been able to move past that point.

  • 1
    $\begingroup$ Eliminating one variable leaves a quartic in the other. There's an explicit "quartic formula", so your system does have a closed-form solution ... but it's super-ugly. $\endgroup$
    – Blue
    Mar 28, 2020 at 19:33
  • 2
    $\begingroup$ Alternatively, you can use the algorithm described here for computing the intersection of two conics, but again, there’s not likely to be a “nice” general closed-form solution. $\endgroup$
    – amd
    Mar 28, 2020 at 20:06
  • $\begingroup$ @Blue Checked and confirmed, unfortunately. $\endgroup$
    – TheVal
    Mar 29, 2020 at 20:44
  • $\begingroup$ @amd Unfortunately I confirm that the closed-form solution will be ugly. I will include your comments in my answer. $\endgroup$
    – TheVal
    Mar 29, 2020 at 20:45

1 Answer 1


As indicated in the comments, there is no "clean" closed-form solution, as a quartic equation has to be eventually solved, either with direct methods or particular algorithms.

A proposed solution method exploits the property of the first equation $f(\xi)=g(\eta)$, since two functions of two separate variables are equated, both must be equal to a constant unique value $K$.

The first equation (RHS) becomes: $$ (C+\eta)(D+\eta)=K $$ which yields (since $\eta>0$) $$ \eta(K)=\frac{-(C+D)+\sqrt{(C+D)^2-4(CD-K)}}{2} \ $$ The first equation (LHS) becomes: $$ (1-\xi)(B+\xi)=K/A $$ which yields (since $\xi>0$) $$ \xi(K)=\frac{-(B-1)+\sqrt{(B-1)^2-4(K/A-B)}}{2} \ $$ The second equation becomes: $$ (F+[\xi-\eta])(G+[\xi-\eta])=EK/A $$ which yields (since $\xi-\eta>0$) $$ [\xi-\eta](K)=\frac{-(F+G)+\sqrt{(F+G)^2-4(FG-EK/A)}}{2} \ $$ Then, the unique constant $K$ must satisfy $$ [\xi-\eta](K)=\xi(K)-\eta(K)\,, $$ thus by substituting the respective functional form of the above functions, the unique constant $K$ is obtained through an irrational equation.


In the special case when $C=D=F=G=0$, the solutions become: $$ \eta(K)=\sqrt{K} $$ and $$ [\xi-\eta](K)=\sqrt{\frac{EK}{A}} \ $$ Thus, by combining the above solutions, $\xi$ is obtained. Since $$ \xi(K)=\eta(K)\left(1+\sqrt{\frac{E}{A}}\right)=\sqrt{K}\left(1+\sqrt{\frac{E}{A}}\right) \,, $$ then by using the previous definition of $\xi(K)$, that is $$ \xi(K)=\sqrt{K}\left(1+\sqrt{\frac{E}{A}}\right)=\frac{-(B-1)+\sqrt{(B-1)^2-4(K/A-B)}}{2} \,, $$ becomes an irrational equation with closed form solution, being: $$ \sqrt{K}=\sqrt{A}\frac{-\alpha\beta+\sqrt{(\alpha\beta)^2+4B(\alpha^2+4)}}{\alpha^2+4} $$ where $\alpha=2(\sqrt{A}+\sqrt{E})$ and $\beta=B-1$. In this special case a quartic has been reduced to two decoupled quadratics.


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