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Difference between revisions of "2005 AMC 12A Problems/Problem 9"

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== Problem ==
 
== Problem ==
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There are two values of <math>a</math> for which the equation <math>4x^2 + ax + 8x + 9 = 0</math> has only one solution for <math>x</math>. What is the sum of these values of <math>a</math>?
+
There are two values of <math>a</math> for which the equation <math>4x^2 + ax + 8x + 9 = 0</math> has only one solution for <math>x</math>. What is the sum of those values of <math>a</math>?
  
 
<math>(\mathrm {A}) \ -16 \qquad (\mathrm {B}) \ -8 \qquad (\mathrm {C})\ 0 \qquad (\mathrm {D}) \ 8 \qquad (\mathrm {E})\ 20</math>
 
<math>(\mathrm {A}) \ -16 \qquad (\mathrm {B}) \ -8 \qquad (\mathrm {C})\ 0 \qquad (\mathrm {D}) \ 8 \qquad (\mathrm {E})\ 20</math>
  
 
== Solution ==
 
== Solution ==
−
=== Video Solution ===
+
=== Video Solution by OmegaLearn===
  
 
https://youtu.be/3dfbWzOfJAI?t=222
 
https://youtu.be/3dfbWzOfJAI?t=222
−
~pi_is_3.14
+
~AVM2023
  
 
=== Solution 1 (Slowest)===
 
=== Solution 1 (Slowest)===
 
We first rewrite <math>4x^2 + ax + 8x + 9 = 0</math> as <math>4x^2 + (a+8)x + 9 = 0</math>. Since there is only one root, the discriminant, <math>(a+8)^2-144</math>, has to be 0. We solve the remaining as below:
 
We first rewrite <math>4x^2 + ax + 8x + 9 = 0</math> as <math>4x^2 + (a+8)x + 9 = 0</math>. Since there is only one root, the discriminant, <math>(a+8)^2-144</math>, has to be 0. We solve the remaining as below:
 
<math>(a+8)^2 = a^2+16a+64</math>.
 
<math>(a+8)^2 = a^2+16a+64</math>.
−
<math>a^2+16a+64-144=a^2+16a-80</math>.
+
<math>a^2+16a+64-144=a^2+16a-80=0</math>.
 
To apply the quadratic formula, we rewrite <math>a^2+16a-80</math> as <math>a^2+16a+-80</math>.
 
To apply the quadratic formula, we rewrite <math>a^2+16a-80</math> as <math>a^2+16a+-80</math>.
 
Then the formula yields:
 
Then the formula yields:
Line 19: Line 19:
 
Which is,
 
Which is,
 
<math>-8\pm12</math>.
 
<math>-8\pm12</math>.
−
This gives <math>-20</math> and <math>4</math>, which sums up to <math>\boxed{-16\text{ (A)}}</math>
+
This gives <math>-20</math> and <math>4</math>, which sums up to <math>\boxed{-16\text{ (A)}}</math>.
 
~AVM2023
 
~AVM2023
  
 
=== Solution 2 (Slow)===
 
=== Solution 2 (Slow)===
−
We first rewrite <math>4x^2 + ax + 8x + 9 = 0</math> as <math>4x^2 + (a+8)x + 9 = 0</math>. Since there is only one root, the discriminant, <math>(a+8)^2-144</math>, has to be 0. We solve the remaining as below:
+
We first rewrite <math>4x^2 + ax + 8x + 9 = 0</math> as <math>4x^2 + (a+8)x + 9 = 0</math>. Since there is only one root, the discriminant, <math>(a+8)^2-144</math>, has to be 0. We expand <math>(a+8)^2</math> as <math>a^2+16a+64</math>. Applying our discriminant rule yields:
−
<math>(a+8)^2 = a^2+16a+64</math>.
+
<math>a^2+16a+64-144=a^2+16a-80=0</math>.
−
<math>a^2+16a+64-144=a^2+16a-80</math>.
 
 
To apply the quadratic formula, we rewrite <math>a^2+16a-80</math> as <math>a^2+16a+-80</math>.
 
To apply the quadratic formula, we rewrite <math>a^2+16a-80</math> as <math>a^2+16a+-80</math>.
 
Then the formula yields:
 
Then the formula yields:
Line 31: Line 30:
 
Which is,
 
Which is,
 
<math>-8\pm12</math>.
 
<math>-8\pm12</math>.
−
Notice that we have to find the sum of the two values, since the average is obviously <math>-8</math>, the sum is <math>2\cdot-8=\boxed{-16\text{ (A)}}</math>
+
Notice that we have to find the sum of the two values, since the average is obviously <math>-8</math>, the sum is <math>2\cdot-8=\boxed{-16\text{ (A)}}</math>.
 
~AVM2023
 
~AVM2023
  
−
=== Solution 3 ===
+
=== Solution 3 (Quick)===
−
Using the [[discriminant]], the result must equal <math>0</math>.
+
We first rewrite <math>4x^2 + ax + 8x + 9 = 0</math> as <math>4x^2 + (a+8)x + 9 = 0</math>. Since there is only one root, the discriminant, <math>(a+8)^2-144</math>, has to be 0. We solve the remaining as below:
−
<math>D = b^2 - 4ac</math>
+
<math>(a+8)^2-144=0</math> so
−
<math> = (a+8)^2 - 4(4)(9)</math>
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<math>(a+8)^2=144</math>.
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<math> = a^2 + 16a + 64 - 144</math>
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So <math>a+8</math> is either <math>12</math> or <math>-12</math>, which make <math>a</math> either <math>4</math> or <math>-20</math>, respectively. The sum of these values is <math>\boxed{-16\text{ (A)}}</math>.
−
<math> = a^2 + 16a - 80 = 0 \Rightarrow</math>
+
~AVM2023
−
<math>   (a + 20)(a - 4) = 0</math>
 
−
Therefore, <math>a = -20</math> or <math>a = 4</math>, giving a sum of <math>-16 \Rightarrow \mathrm{ (A)}</math>.
 
  
−
===Solution 4===
+
=== Solution 3 (Quickest)===
−
First, notice that for there to be only <math>1</math> 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 <math>a</math> must be such that both <math>(2x+3)^2</math> and <math>(2x-3)^2</math>. Clearly, <math>a=4</math> or <math>a=-20</math>. Hence <math>-20 + 4 = -16 \Rightarrow \mathrm{(A)}</math>.
+
We first rewrite <math>4x^2 + ax + 8x + 9 = 0</math> as <math>4x^2 + (a+8)x + 9 = 0</math>. Since there is only one root, the discriminant, <math>(a+8)^2-144</math>, has to be 0. We solve the remaining as below:
−
 
+
<math>(a+8)^2-144=0</math> so
−
Solution by franzliszt
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<math>(a+8)^2=144</math>.
 +
So <math>a+8</math> is either <math>12</math> or <math>-12</math>, which make <math>2a+16=0</math>, the sum, or <math>2a</math>, is <math>\boxed{-16\text{ (A)}}</math>.
 +
~AVM2023
  
 
== See also ==
 
== See also ==

Latest revision as of 14:49, 1 July 2025

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 those 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 (Quickest)

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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