2003 AIME II Problems/Problem 9
Contents
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
Solution 3
So we just have to find: .
And by Newton's Sums this computes to: .
~ LuisFonseca123
Solution 4
If we scale by , we get . In order to get to , we add . Therefore, our answer is . However, rearranging , makes our final answer . The sum of the squares of the roots is and the sum of the roots is . Adding 4 to our sum, we get .
~ Vedoral
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 |
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