4. Given the equation about $x$: $k x^{2}+\frac{|x|}{x+6}=0$ has four distinct real solutions. Then the range of the real number $k$ is $\qquad$ .
4. $k<-\frac{1}{9}$.

Obviously, $k \neq 0, x=0$ is a known solution of the equation.
When $x \neq 0$, the condition becomes $|x|(x+6)=-\frac{1}{k}$ having
three distinct non-zero real solutions.
From the graph of $f(x)=|x|(x+6)$ (as shown in Figure 6), we get $0<-\frac{1}{k}<f(-3) \Rightarrow 0<-\frac{1}{k}<9$
$\Rightarrow k<-\frac{1}{9}$.