imc

A quadratic equation has exactly one root if and only if it is a perfect square. So set $4x^2+ax+8x+9=(mx+n{)}^2$ $4x^2+ax+8x+9=m^2{x}^2+2mnx+n^2$ Two polynomials are equal only if their coefficients are equal, so we must have $m^2=4,n^2=9$ $m=\pm 2,n=\pm 3$ $a+8=2mn=\pm 2\cdot 2\cdot 3=\pm 12$ $a=4$ or $a=-20$. So the desired sum is $(4)+(-20)=\boxed{\text{(A)}-16}$ Alternatively, note that whatever the two values of $a$ are, they must lead to equations of the form $p{x}^2+qx+r=0$ and $p{x}^2-qx+r=0$. So the two choices of $a$ must make $a_1+8=q$ and $a_2+8=-q$ so $a_1+a_2+16=0$ and $a_1+a_2=\boxed{-16}$