2005 AMC 10A Problem 10

The equation has two $x$ terms, so first combine them. Then use the rule for when a quadratic has just one solution.

1. Combine like terms. $ax + 8x = (a + 8)x$, so the equation is
$$4x^2 + (a + 8)x + 9 = 0.$$
2. One solution means the discriminant is zero. For $Ax^2 + Bx + C = 0$ there is exactly one solution when $B^2 - 4AC = 0$. Here $A = 4$, $B = a + 8$, $C = 9$:
$$(a + 8)^2 - 4 \cdot 4 \cdot 9 = 0, \qquad\text{so}\qquad (a + 8)^2 = 144.$$
3. Solve. A number whose square is $144$ is $12$ or $-12$.
- $a + 8 = 12$ gives $a = 4$.
- $a + 8 = -12$ gives $a = -20$.
4. Add them. $4 + (-20) = -16$.

Check with $a = 4$: the equation is $4x^2 + 12x + 9 = (2x + 3)^2 = 0$, which has the single solution $x = -\frac{3}{2}$.

The answer is $\boxed{\textbf{(A)}\ -16}$.

A quadratic with only one solution is a perfect square, and that happens exactly when the discriminant is zero. The key is to use the whole middle coefficient, $a + 8$. The two answers are $-8 + 12$ and $-8 - 12$, balanced around $-8$, so their sum is $2 \times (-8) = -16$ without even finding them.

First collect like terms: the equation is $4x^2 + (a+8)x + 9 = 0$.

A quadratic has exactly one solution when its discriminant is zero:
$$(a+8)^2 - 4 \cdot 4 \cdot 9 = 0 \quad\Longrightarrow\quad (a+8)^2 = 144 \quad\Longrightarrow\quad a + 8 = \pm 12.$$
So $a = 4$ or $a = -20$, and their sum is $4 + (-20) = -16$.

The answer is $\boxed{\textbf{(A)}\ -16}$.