equation involving absolute value function has $4$ solutions 
If the equation $|x^2-5|x|+k|-\lambda x+7 \lambda = 0$ has exactly $4$ solution, Then $(\lambda,k)$ is 

Attempt:$x^2-5|x|+k = \lambda x-7 \lambda$
$\star$ For $x>0,$ then $x^2-5x+k=\lambda x- 7 \lambda.$
$\star$ For $x\leq 0,$ then $x^2+5x+k=\lambda x- 7 \lambda.$
could some help me how to go further,thanks

First, notice that if $\lambda=k=0$ then the only solutions are $0,\,-5$ and $5$.
Suppose $\lambda=0$ and $0<k<\dfrac{25}{4}$.
Then
\begin{eqnarray}
 {\Big\vert}\, |x|^2-5|x|+k\, {\Big\vert}&=&0\\
{\Huge\vert}\left(|x|-\frac{5}{2}\right)^2+k-\frac{25}{4}{\Huge\vert}&=&0\\
\left(|x|-\frac{5}{2}\right)^2&=&\frac{25}{4}-k\\
|x|&=&\frac{5}{2}\pm\frac{5}{2}\sqrt{1-\frac{4k}{25}}>0
\end{eqnarray}
Thus there are only four solutions
$$ x=\pm\frac{5}{2}\left(1\pm\sqrt{1-\frac{4k}{25}}\,\right) $$
provided that $(\lambda,k)=(0,k)$ and $0<k<\dfrac{25}{4}$.
This does not exhaust the solutions. Graphing techniques indicate that if $k=0$ then there is an open interval containing $\lambda=-1$ for which there are only four real solutions. But perhaps we are including complex solutions?
Was the object to characterize all possible solutions?
