Difference between revisions of "2005 AMC 12A Problems/Problem 9"
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== Problem == | == Problem == | ||
− | 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 | + | 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> |
Latest revision as of 14:49, 1 July 2025
Contents
Problem
There are two values of for which the equation
has only one solution for
. What is the sum of those values of
?
Solution
Video Solution by OmegaLearn
https://youtu.be/3dfbWzOfJAI?t=222 ~AVM2023
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:
.
.
To apply the quadratic formula, we rewrite
as
.
Then the formula yields:
.
Which is,
.
This gives
and
, which sums up to
.
~AVM2023
Solution 2 (Slow)
We first rewrite as
. Since there is only one root, the discriminant,
, has to be 0. We expand
as
. Applying our discriminant rule yields:
.
To apply the quadratic formula, we rewrite
as
.
Then the formula yields:
.
Which is,
.
Notice that we have to find the sum of the two values, since the average is obviously
, the sum is
.
~AVM2023
Solution 3 (Quick)
We first rewrite as
. Since there is only one root, the discriminant,
, has to be 0. We solve the remaining as below:
so
.
So
is either
or
, which make
either
or
, respectively. The sum of these values is
.
~AVM2023
Solution 3 (Quickest)
We first rewrite as
. Since there is only one root, the discriminant,
, has to be 0. We solve the remaining as below:
so
.
So
is either
or
, which make
, the sum, or
, is
.
~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 |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.