There are two values of $a$ for which the equation $4x^2 + ax + 8x + 9 = 0$ has only one solution for $x$. What is the sum of those values of $a$?
$\textbf{(A) }-16\qquad\textbf{(B) }-8\qquad\textbf{(C) } 0\qquad\textbf{(D) }8\qquad\textbf{(E) }20$
A [quadratic equation](https://artofproblemsolving.com/wiki/index.php/Quadratic_equation) has exactly one [root](https://artofproblemsolving.com/wiki/index.php/Root) if and only if it is a [perfect square](https://artofproblemsolving.com/wiki/index.php/Perfect_square).  So set
$4x^2 + ax + 8x + 9 = (mx + n)^2$
$4x^2 + ax + 8x + 9 = m^2x^2 + 2mnx + n^2$
Two [polynomials](https://artofproblemsolving.com/wiki/index.php/Polynomial) are equal only if their [coefficients](https://artofproblemsolving.com/wiki/index.php/Coefficient) 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 $(4)+(-20)=\boxed{\textbf{(A)}-16}$

Alternatively, note that whatever the two values of $a$ are, they must lead to equations of the form $px^2 + qx + r =0$ and $px^2 - qx + r = 0$.  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 $a_1 + a_2 =\boxed{\textbf{(A)}-16}$