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2005 AMC 12A Problems/Problem 9

Revision as of 11:15, 30 December 2024 by Avm2023 (talk | contribs) (→‎Video Solution by OmegaLearn)

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 ~AVM2023

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 expand $(a+8)^2$ as $a^2+16a+64$. Applying our discriminant rule yields: $a^2+16a+64-144=a^2+16a-80=0$. 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: $(a+8)^2-144=0$ so $(a+8)^2=144$. 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 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: $(a+8)^2-144=0$ so $(a+8)^2=144$. So $a+8$ is either $12$ or $-12$, which make $2a+16=0$, the sum, or $2a$, is $\boxed{-16\text{ (A)}}$. ~AVM2023

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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