Amount of solution pairs $(e,f)$ of this system of equations?

Question

I search every pair $(e,f)$ of real numbers ($\,\Bbb R\,$) that fulfills this system of equations:

$$ \begin{align} e^2 + d \cdot f = d \\ e + d^2 \cdot f^2 = d \end{align} $$

I have to calculate for

a) d=1 every pair (e,f) and

b) i want to know for any real number $d$ , the amount of different solution pairs $(e,f)$.

Is the quartic formula really necessary or can I use the other solution?

Answer 1

For a given $d$ you have two equations in two unknowns. You can do $$e^2+df=d\\e+d^2f^2=d\\f=\frac {e^2}d-1\\e+d^2\left(\frac {e^2}d-1\right)^2=d\\e^4-2e^2d+e+d^2-d=0$$ which is a quartic in $e$. There is a very messy formula for that or you can solve it numerically.

Answer 2

From the first equation we get by squaring: $$d^2f^2=(d-e^2)^2$$ plugging this in the second one we get $$e+d^2-2de^2+e^4=d$$

Unfortunately we get a polynom of degree four. Good luck!

Answer 3

Hint:   subtracting the two equations gives:

$$ \begin{align} e^2 - d^2 \cdot f^2 - e + d \cdot f = 0 \;\;&\iff\;\; (e-df)(e+df)-(e-df) = 0 \\ &\iff\;\; (e-df)(e+df-1)=0 \end{align} $$

For the latter to hold, (at least) one of the factors must be $\,0\,$, so either:

• $e-df=0\,$, then substituting $\,df=e\,$ in the first equation gives $\,e^2+e-d=0\,$;

• $e+df-1=0\,$, then substituting $\,df=1-e\,$ back gives $\,e^2-e+1-d=0\,$.

It follows that there are at most $\,4\,$ solutions for $\,e\,$ in general, and each eligible one corresponds univocally to an $\,f\,$ given by the respective formula (assuming $\,d \ne 0\,$ of course).