Difference between revisions of "2003 AIME II Problems/Problem 9"
m (→Solution) |
Sleepypuppy (talk | contribs) (→Video Solution by Sal Khan) |
||
Line 41: | Line 41: | ||
https://www.youtube.com/watch?v=ZSESJ8TeGSI&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=14 | https://www.youtube.com/watch?v=ZSESJ8TeGSI&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=14 | ||
- AMBRIGGS | - AMBRIGGS | ||
+ | Nice! | ||
== See also == | == See also == |
Revision as of 15:57, 12 December 2022
Problem
Consider the polynomials and Given that and are the roots of find
Solution
When we use long division to divide by , the remainder is .
So, since is a root, .
Now this also follows for all roots of Now
Now by Vieta's we know that , so by Newton's Sums we can find
So finally
Solution 2
Let then by Vieta's Formula we have By Newton's Sums we have
Applying the formula couples of times yields .
~ Nafer
Video Solution by Sal Khan
https://www.youtube.com/watch?v=ZSESJ8TeGSI&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=14 - AMBRIGGS Nice!
See also
2003 AIME II (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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.