Difference between revisions of "2024 AMC 12A Problems/Problem 15"
m |
(→Solution 1) |
||
Line 10: | Line 10: | ||
-ev2028 | -ev2028 | ||
+ | ==Solution 3== | ||
+ | First, denote that | ||
+ | <cmath>p+q+r=-2, | ||
+ | pq+pr+qr=-1, | ||
+ | pqr=-3</cmath> | ||
+ | Then we expand the expression | ||
+ | <cmath>(p^2+4)(q^2+4)(r^2+4)</cmath> | ||
+ | <cmath>=(pqr)^2+4((pq)^2+(pr)^2+(qr)^2)+4^2(p^2+q^2+r^2)+4^3</cmath> | ||
+ | <cmath>=(-3)^2+4((pq+pr+qr)^2-2pqr(p+q+r))+4^2((p+q+r)^2-2(pq+pr+qr))+4^3</cmath> | ||
+ | <cmath>=(-3)^2+4((-1)^2-2(-3)(-2))+4^2((-2)^2-2(-1))+4^3</cmath> | ||
+ | <cmath>=\fbox{(D) 125}</cmath> | ||
==Solution 2== | ==Solution 2== |
Revision as of 18:05, 8 November 2024
Contents
[hide]Problem
The roots of are and What is the value of
Solution 1
You can factor as (p−2i)(p+2i)(q−2i)(q+2i)(r−2i)(r+2i).
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
Solution 3
First, denote that Then we expand the expression
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.