<s>[INST] 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
 [/INST] 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) $x^2-  5x +k = -\lambda x+7 \lambda\implies x^2 + x(-5 + \lambda) + (k-7\lambda) = 0$
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