Difference between revisions of "2003 AIME II Problems/Problem 9"
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Now <cmath>P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1</cmath> | Now <cmath>P(z_2)+P(z_1)+P(z_3)+P(z_4)=z_1^2-z_1+1+z_2^2-z_2+1+z_3^2-z_3+1+z_4^2-z_4+1</cmath> | ||
− | Now by Vieta's we know that <math>-z_4-z_3-z_2-z_1=-1</math>, | + | Now by [[Vieta's formulas|Vieta's]] we know that <math>-z_4-z_3-z_2-z_1=-1</math>, |
so by [[Newton's Sums]] we can find <math>z_1^2+z_2^2+z_3^2+z_4^2</math> | so by [[Newton's Sums]] we can find <math>z_1^2+z_2^2+z_3^2+z_4^2</math> | ||
Revision as of 18:19, 10 January 2024
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
[rule]
Nice!-sleepypuppy
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.