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2005 AMC 10A Problem 10

2005 AMC 10A Problem 10AMC 10 Step by Step

Name: Date: September 13, 2026

Quadratics · difficulty 2 of 5 · about 2 min · position P1-10

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 $a$ ?

A)
$-16$
B)
$-8$
C)
$0$
D)
$8$
E)
$20$

Answer

Key insight

Combine the x-terms into (a+8)x; one solution means discriminant zero, so a + 8 = plus or minus 12.

Solution

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

Why this works

"Only one solution" for a quadratic is the discriminant condition $b^2 = 4ac$ , and here $b$ is the combined coefficient $a + 8$ . The two values of $a$ are symmetric around $-8$ (they are $-8 \pm 12$ ), which is why their sum is $2(-8) = -16$ without ever computing $\sqrt{144}$ .

Alternative approach

Expand the discriminant equation into $a^2 + 16a - 80 = 0$ ; by Vieta the two values of $a$ sum to $-16$ .

The trap

Forgetting to merge ax and 8x first, so the discriminant is written as a^2 - 144 and the sum comes out 0.

Common mistakes

Forgetting to merge ax and 8x first, so the discriminant is written as a^2 - 144 and the sum comes out 0.
Taking only the positive square root, finding $a = 4$ alone, and then having nothing to sum.

Techniques

Apply an identity: SFFT, sum of squares, difference of cubes, Vieta