Difference between revisions of "2024 AMC 12A Problems/Problem 15"
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Revision as of 23:04, 9 November 2024
Contents
[hide]Problem
The roots of are
and
What is the value of
Solution 1
You can factor as
.
For any polynomial , you can create a new polynomial
, which will have roots that instead have the value subtracted.
Substituting and
into
for the first polynomial, gives you
and
as
for both equations. Multiplying
and
together gives you
.
-ev2028
~Latex by eevee9406
Solution 2
Let . Then
.
We find that and
, so
.
~eevee9406
Solution 3
First, denote that
Then we expand the expression
~lptoggled
Solution 4 (Reduction of power)
The motivation for this solution is the observation that is easy to compute for any constant c, since
(*), where
is the polynomial given in the problem. The idea is to transform the expression involving
into one involving
.
Since is a root of
,
which gives us that
. Then
Since
and
are also roots of
, the same analysis holds, so
(*) This is because
since
for all
.
~tsun26 ~KSH31415 (final step and clarification)
See also
2024 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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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