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 by OmegaLearn

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

Solution 1 (Slowest)

We first rewrite $4x^2 + ax + 8x + 9 = 0$ as $4x^2 + (a+8)x + 9 = 0$ . Since there is only one root, the discriminant, $(a+8)^2-144$ , has to be 0. We solve the remaining as below: $(a+8)^2 = a^2+16a+64$ . $a^2+16a+64-144=a^2+16a-80=0$ . To apply the quadratic formula, we rewrite $a^2+16a-80$ as $a^2+16a+-80$ . Then the formula yields: $\frac{-16\pm24}{2}$ . Which is, $-8\pm12$ . This gives $-20$ and $4$ , which sums up to $\boxed{-16\text{ (A)}}$ . ~AVM2023

Solution 2 (Slow)

We first rewrite $4x^2 + ax + 8x + 9 = 0$ as $4x^2 + (a+8)x + 9 = 0$ . Since there is only one root, the discriminant, $(a+8)^2-144$ , has to be 0. We solve the remaining as below: $(a+8)^2 = a^2+16a+64$ . $a^2+16a+64-144=a^2+16a-80=0$ . To apply the quadratic formula, we rewrite $a^2+16a-80$ as $a^2+16a+-80$ . Then the formula yields: $\frac{-16\pm24}{2}$ . Which is, $-8\pm12$ . Notice that we have to find the sum of the two values, since the average is obviously $-8$ , the sum is $2\cdot-8=\boxed{-16\text{ (A)}}$ . ~AVM2023

Solution 3 (Quick)

We first rewrite $4x^2 + ax + 8x + 9 = 0$ as $4x^2 + (a+8)x + 9 = 0$ . Since there is only one root, the discriminant, $(a+8)^2-144$ , has to be 0. We solve the remaining as below: $(a+8)^2-144=0$ . $(a+8)^2=144$ . So $a+8$ is either $12$ or $-12$ , which make $a$ either $4$ or $-20$ , respectively. The sum of these values is $\boxed{-16\text{ (A)}}$ . ~AVM2023

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

2005 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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