jmc

The square of the binomial $rx+s$ is $$(rx+s{)}^2=r^2{x}^2+2rsx+s^2.$$If this is equal to $a{x}^2+12x+9$, then $s$ must be either 3 or $-3$. Since $(rx+s{)}^2=(-rx-s{)}^2$, we may choose either $s=3$ or $s=-3$, and the solution will be the same. We choose $s=3$.

The square of $rx+3$ is $$(rx+3{)}^2=r^2{x}^2+6rx+9.$$If this is equal to $a{x}^2+12x+9$ then we must have $12=6r$ or $r=2$. This gives our square: $${\left(2x+3\right)}^2=4x^2+12x+9.$$Therefore $a=\boxed{4}$.