How can you find $m$ in $mx^2+(m-3)x+1=0 $ so that there is only one solution How can you find $m$ in $$mx^2+(m-3)x+1=0 $$ so that there is only one solution.
I tried to solve it by quadratic equation but I end up with two solutions.
So I want it know that is there a way so that I'll only get one solution the end.
Thanks.

Let $x=\alpha$ be one solution of the given quadratic equation: $mx^2+(m-3)x+1=0$ then $\alpha$ & $\alpha$ be the roots of the given equation $$\text{sum of roots}=\alpha+\alpha=-\frac{m-3}{m}$$$$\alpha=\frac{3-m}{2m}\tag 1$$
$$\text{product of roots}=\alpha\cdot \alpha=\frac{1}{m}$$$$\alpha^2=\frac{1}{m}\tag 2$$
substituting the value of $\alpha$ from (2) into (1), one should get 
$$\left(\frac{3-m}{2m}\right)^2=\frac 1m$$
$$m^2-10m+9=0$$
using quadratic formula, 
$$m=\frac{-(-10)\pm\sqrt{(-10)^2-4(1)(9)}}{2(1)}=5\pm4$$
Thus, the given quadratic equation: $mx^2+(m-3)x+1=0$ will have only one solution for $\color{red}{m=9}\ \text{or} $ $ \ \ \color{red}{m=1}$. 
