2024 AMC 12A Problems/Problem 15
Contents
Problem
The roots of are and What is the value of
Solution 1
You can factor (p^2 + 4)(q^2 + 4)(r^2 + 4) as (p − 2i)(p + 2i)(q − 2i)(q + 2i)(r − 2i)(r + 2i).
For any polynomial f(x), you can create a new polynomial f(x+2), which will have roots that instead have the value subtracted.
Substituting x-2 and x+2 into x for the first polynomial, gives you 10i-5 and -10i-5 as c for both equations. Multiplying 10i-5 and -10i-5 together gives you 125
-ev2028
Solution 2
Let . Then .
We find that and , so .
~eevee9406
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.