Number of rational roots of the equation $\left|{\mathrm{x}}^2-2\mathrm{x}-3\right|+4\mathrm{x}=0$ is

$$\begin{gathered} \left|x^2-2x-3\right|+4x=0 \\ \Rightarrow \left|\left(\mathrm{x}+1\right)\left((\mathrm{x}-3\right)\right|+4\mathrm{x}=0 \end{gathered}$$
$$\begin{gathered} case\left(i\right) x^2-2x-3>0 \text{For }\mathrm{x}\in \left(-\mathrm{\infty },-1]\cup [3,\mathrm{\infty }\right) \\ \begin{array}{l}{x}^2-2x-3+4x=0\\ \Rightarrow x^2+2x-3=0\\ \Rightarrow (x+3)(x-1)=0\\ \Rightarrow x=-3\text{ and }x=1\\ \text{ However }\mathrm{x}=1\text{ does not fall in the above range. }\\ \text{ Hence }x=-3\text{ is a possible real root. }\\ \text{case}\left(\mathrm{ii}\right)x^2-2x-3<0 \text{ For }x\in \left(-1,3\right)\\ \begin{array}{l}-x^2+2x+3+4x=0\\ \Rightarrow -x^2+6x+3=0\\ \Rightarrow x^2-6x-3=0\\ \Rightarrow (x-3{)}^2-12=0\\ \Rightarrow (x-3{)}^2=12\\ \Rightarrow (x-3)=\pm 2\sqrt{3}\\ x=3\pm 2\sqrt{3}\dots \text{ are irrational roots. }\\ \text{ Hence there exists only one rational root that is }\mathrm{x}=-3\end{array}\end{array} \end{gathered}$$