If $\frac{px+q}{{(2x-3)}^2}=\frac{1}{2x-3}+\frac{3}{{(2x-3)}^2}$ then $p,q$ are roots of quadratic equation is

$\frac{px+q}{{(2x-3)}^2}=\frac{1(2x-3)+3}{{(2x-3)}^2}$

$px+q=2x-3+3$

$px+q=2x$

Comparing coefficients, $p=2,q=0$

Now $\Rightarrow p+q=2\text{\hspace{0.17em}and\hspace{0.17em}}pq=0$

$\therefore$ Quadratic equation $\Rightarrow k[x^2-(p+q)x+pq]=0$

$k[x^2-2x+0]=0$

If $k=1$, then the equation is

$\therefore x^2-2x=0$