Difference between revisions of "2005 AMC 12A Problems/Problem 9"
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=== 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)x)^2-144</math>, has to be 0. We solve the remaining as below: | |
=== Solution 2 === | === Solution 2 === | ||
Revision as of 10:47, 30 December 2024
Contents
Problem
There are two values of 
for which the equation 
has only one solution for 
. What is the sum of these values of ?
Solution
Video Solution
https://youtu.be/3dfbWzOfJAI?t=222 ~pi_is_3.14
Solution 1 (Slowest)
We first rewrite as 
. Since there is only one root, the discriminant, 
, has to be 0. We solve the remaining as below:
Solution 2
Another method would be to use the quadratic formula, since our 
coefficient is given as 4, the coefficient is 
and the constant term is 
. Hence, 
Because we want only a single solution for , the determinant must equal 0. Therefore, we can write 
which factors to 
; using Vieta's formulas we see that the sum of the solutions for is the opposite of the coefficient of , or 
.
Solution 3
Using the discriminant, the result must equal 
.
Therefore, 
or 
, giving a sum of .
Solution 4
First, notice that for there to be only 
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 must be such that both 
and 
. Clearly, 
or 
. Hence 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
