Solving a system of equations involving an absolute value

Question

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:

$$ \begin{cases} \text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\epsilon^2=\epsilon\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right)\\ \\ \epsilon=\pm\frac{\text{c}}{\text{d}} \end{cases}\to\text{a}\cdot\text{c}+\text{b}\cdot\text{d}\cdot\frac{\text{c}^2}{\text{d}^2}=\pm\frac{\text{c}}{\text{d}}\cdot\left(\text{b}\cdot\text{c}-\text{a}\cdot\text{d}\right) $$

Answer 1

A case disctinction helps. Assume first that $\epsilon=c/d$. Then we obtain $ac+bc^2/d=c^2b/d-ca$, hence $2ac=0$. Both cases $a=0$ or $c=0$ are easily solved.

Answer 2

If $c/d$ is positive, you will have $|-c/d|=c/d$. Thus

$$ ac+bd\frac{c^2}{d^2}=\dfrac{c}{d}(bc-ad), $$

or

$$ ac+\frac{bc^2}{d}=\dfrac{bc^2}{d}-ac, $$

or

$$ac=0.$$

Now if $c/d$ is negative, you will have $|-c/d|=-c/d$. Thus

$$ ac+bd\frac{c^2}{d^2}=-\dfrac{c}{d}(bc-ad), $$

or

$$ ac+\frac{bc^2}{d}=-\dfrac{bc^2}{d}+ac, $$

or

$$bc^2/d=0.$$

Answer 3

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: $$ \begin{gathered} \left\{ \begin{gathered} bd\varepsilon ^{\,2} - \left( {bc - ad} \right)\varepsilon + ac = 0 \hfill \\ \varepsilon = \pm \frac{c} {d} \hfill \\ \end{gathered} \right.\quad \Rightarrow \quad \left\{ \begin{gathered} bc - ad = 0 \hfill \\ b\frac{{c^{\,2} }} {d} + ac = 0 \hfill \\ \end{gathered} \right.\quad \Rightarrow \hfill \\ \Rightarrow \quad \left\{ \begin{gathered} bc - ad = 0 \hfill \\ d \ne 0 \hfill \\ \left( {bc + ad} \right)c = 0 \hfill \\ \end{gathered} \right. \Rightarrow \quad \left\{ \begin{gathered} c = 0 \hfill \\ d \ne 0 \hfill \\ a = 0 \hfill \\ \forall b \hfill \\ \end{gathered} \right. \hfill \\ \end{gathered} $$