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
 A: 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?
A: No harm in doing it in a case by case basis.  
You have 
$x^2\pm 5x +k = \pm(\lambda x-7 \lambda)$
That's four equations to solve:
1) $x^2+  5x +k = \lambda x-7 \lambda\implies x^2 + x(5 - \lambda) + (k+7\lambda) = 0$
2) $x^2- 5x +k = \lambda x-7 \lambda \implies x^2 + x(-5 - \lambda) + (k+7\lambda) = 0$
3) $x^2+  5x +k = -\lambda x+7 \lambda\implies x^2 + x(5 + \lambda) + (k-7\lambda) = 0$
4) $x^2-  5x +k = -\lambda x+7 \lambda\implies x^2 + x(-5 + \lambda) + (k-7\lambda) = 0$
A: I arranged a graph for you: https://www.desmos.com/calculator/fdrcwuc5gg
Try to use the slider you will find the following cases:
$k<0$) 2 sets of solutions: ${k \over 7}<\lambda <0$ and $-19+2\sqrt{k+84}<\lambda <-9+2\sqrt{k+14}$
$0<k<231-14\sqrt{266}$) 4 intersections for: $-19+2\sqrt{k+84}<\lambda <-9+2\sqrt{k+14}$
$231-14\sqrt{266}<k<161-42\sqrt{14}$) 4 intersections for: $-9+2\sqrt{k+14}<\lambda <-{k\over 7}$
$161-42\sqrt{14}<k<{25\over 4}$) 4 intersections for: $-{k\over 7} <\lambda<-9+2\sqrt{k+14} $
$k>{25\over 4}$) 4 intersections for: $9-2\sqrt{k+14}  <\lambda<-{k\over 7}  $
${25\over 4}$ is the vertex of the parabolas, you'll find the other two numbers equating the lambdas at the extreme points when the line passing through the cusp at $0$ is also tangent to a maximum.
