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2005 AMC 12A Problems/Problem 9

Problem

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 these values of $a$?

$(\mathrm {A}) \ -16 \qquad (\mathrm {B}) \ -8 \qquad (\mathrm {C})\ 0 \qquad (\mathrm {D}) \ 8 \qquad (\mathrm {E})\ 20$

Solution

Video Solution

https://youtu.be/3dfbWzOfJAI?t=222 ~pi_is_3.14

Solution 1

A quadratic equation always has two roots, unless it has a double root. That means we can write the quadratic as a square, and the coefficients 4 and 9 suggest this. Completing the square, $0 = (2x \pm 3)^2 = 4x^2 \pm 12x + 9$, so $\pm 12 = a + 8 \Longrightarrow a = 4, -20$. The sum of these is $-20 + 4 = -16 \Rightarrow \mathrm{(A)}$.

Solution 2

Another method would be to use the quadratic formula, since our $x^2$ coefficient is given as 4, the $x$ coefficient is $a+8$ and the constant term is $9$. Hence, $x = \frac{-(a+8) \pm \sqrt {(a+8)^2-4(4)(9)}}{2(4)}$ Because we want only a single solution for $x$, the determinant must equal 0. Therefore, we can write $(a+8)^2 - 144 = 0$ which factors to $a^2 + 16a - 80 = 0$; using Vieta's formulas we see that the sum of the solutions for $a$ is the opposite of the coefficient of $a$, or $-16 \Rightarrow \mathrm{ (A)}$.

Solution 3

Using the discriminant, the result must equal $0$. $D = b^2 - 4ac$ $= (a+8)^2 - 4(4)(9)$ $= a^2 + 16a + 64 - 144$ $= a^2 + 16a - 80 = 0 \Rightarrow$ $(a + 20)(a - 4) = 0$ Therefore, $a = -20$ or $a = 4$, giving a sum of $-16 \Rightarrow \mathrm{ (A)}$.

Solution 4

First, notice that for there to be only $1$ root to a quadratic, the quadratic must be a square. Then, notice that the quadratic and linear terms are both squares. Thus, the value of $a$ must be such that both $(2x+3)^2$ and $(2x-3)^2$. Clearly, $a=4$ or $a=-20$. Hence $-20 + 4 = -16 \Rightarrow \mathrm{(A)}$.

Solution by franzliszt

See also

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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