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

(→‎Solution 2)
(→‎Solution 1 (Slowest))
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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
  

Revision as of 11:02, 30 December 2024

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 (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: $(a+8)^2 = a^2+16a+64$. $a^2+16a+64-144=a^2+16a-80$. 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: $(a+8)^2 = a^2+16a+64$. $a^2+16a+64-144=a^2+16a-80$. 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

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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