Solving a system of equations involving an absolute value Solve the following system:
$$
\begin{cases}
\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\epsilon^2=\epsilon\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)\\
\\
\left|\epsilon\right|=\left|-\frac{\text{c}}{\text{d}}\right|
\end{cases}
$$
I don't know how to proceed?!

With the help of the comments and the answer I got:
$$
\begin{cases}
\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\epsilon^2=\epsilon\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)\\
\\
\epsilon=\pm\frac{\text{c}}{\text{d}}
\end{cases}\to\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\frac{\text{c}^2}{\text{d}^2}=\pm\frac{\text{c}}{\text{d}}\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)
$$

We are given
$$
\left\{ \begin{gathered}
  bd\varepsilon ^{\,2}  - \left( {bc - ad} \right)\varepsilon  + ac = 0 \hfill \\
  \varepsilon  =  \pm \frac{c}
{d} \hfill \\ 
\end{gathered}  \right.
$$
So we have the equation of a vertical parabola in $\varepsilon$, whose intercepts with the $x$ axis
must be symmetrical (vs. $\varepsilon=0$).
Thus the parabola must be symmetrical as well, i.e. the coefficient
of the $\varepsilon$ term must be null, which gives:
$$
\begin{gathered}
  \left\{ \begin{gathered}
  bd\varepsilon ^{\,2}  - \left( {bc - ad} \right)\varepsilon  + ac = 0 \hfill \\
  \varepsilon  =  \pm \frac{c}
{d} \hfill \\ 
\end{gathered}  \right.\quad  \Rightarrow \quad \left\{ \begin{gathered}
  bc - ad = 0 \hfill \\
  b\frac{{c^{\,2} }}
{d} + ac = 0 \hfill \\ 
\end{gathered}  \right.\quad  \Rightarrow  \hfill \\
   \Rightarrow \quad \left\{ \begin{gathered}
  bc - ad = 0 \hfill \\
  d \ne 0 \hfill \\
  \left( {bc + ad} \right)c = 0 \hfill \\ 
\end{gathered}  \right. \Rightarrow \quad \left\{ \begin{gathered}
  c = 0 \hfill \\
  d \ne 0 \hfill \\
  a = 0 \hfill \\
  \forall b \hfill \\ 
\end{gathered}  \right. \hfill \\ 
\end{gathered} 
$$
