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 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> |
<math>a_ns_2+a_{n-1}s_1+2a_{n-2}=0</math> | <math>a_ns_2+a_{n-1}s_1+2a_{n-2}=0</math> | ||
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So finally | So finally | ||
− | <math>P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{ | + | <math>P(z_2)+P(z_1)+P(z_3)+P(z_4)=3+4-1=\boxed{006}.</math> |
− | |||
== Solution 2 == | == Solution 2 == | ||
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~ Nafer | ~ Nafer | ||
+ | |||
+ | == Solution 3 == | ||
+ | |||
+ | <math>P(x) = x^{2}Q(x)+x^{4}-x^{3}-x.</math> | ||
+ | |||
+ | So we just have to find: <math>\sum_{n=1}^{4} z^{4}_n - \sum_{n=1}^{4} z^{3}_n - \sum_{n=1}^{4} z_n</math>. | ||
+ | |||
+ | And by [[Newton's Sums]] this computes to: <math>11-4-1 = \boxed{006}</math>. | ||
+ | |||
+ | ~ LuisFonseca123 | ||
+ | |||
+ | == Solution 4 == | ||
+ | If we scale <math>Q(x)</math> by <math>x^2</math>, we get <math>x^6-x^5-x^4-x^2</math>. In order to get to <math>P(x)</math>, we add <math>x^4-x^3-x</math>. Therefore, our answer is <math>\sum_{n=1}^{4} z^4_n-z^3_n-z_n</math>. However, rearranging <math>Q(z_n) = 0</math>, makes our final answer <math>\sum_{n=1}^{4} z^2_n-z_n+1</math>. The sum of the squares of the roots is <math>1^2-2(-1) = 3</math> and the sum of the roots is <math>1</math>. Adding 4 to our sum, we get <math>3-1+4 = \boxed{006}</math>. | ||
+ | |||
+ | ~ Vedoral | ||
+ | |||
+ | == Solution 5 == | ||
+ | Let <math>S_k</math> = <math>z_1^k+z_2^k+z_3^k+z_4^k</math> | ||
+ | |||
+ | By [[Newton's Sums]], | ||
+ | |||
+ | <math>S_1-1=0</math> | ||
+ | |||
+ | <math>S_2-S_1-2=0</math> | ||
+ | |||
+ | <math>S_3-S_2-S_1=0</math> | ||
+ | |||
+ | <math>S_4-S_3-S_2-4=0</math> | ||
+ | |||
+ | <math>S_5-S_4-S_3-S_1=0</math> | ||
+ | |||
+ | <math>S_6-S_5-S_4-S_2=0</math> | ||
+ | |||
+ | Solving for <math>S_1,S_2,S_3,S_4,S_5,S_6</math>, we get <math>S_1=1, S_2=3, S_3=4, S_4=11, S_5=16, S_6=30</math> | ||
+ | |||
+ | <math>P(z_1)+P(z_2)+P(z_3)+P(z_4)=S_6-S_5-S_3-S_2-S_1=\boxed{006}</math> | ||
+ | |||
+ | == Video Solution by Sal Khan == | ||
+ | https://www.youtube.com/watch?v=ZSESJ8TeGSI&list=PLSQl0a2vh4HCtW1EiNlfW_YoNAA38D0l4&index=14 | ||
+ | - AMBRIGGS | ||
+ | |||
+ | |||
+ | [rule] | ||
+ | |||
+ | Nice!-sleepypuppy | ||
== See also == | == See also == |
Latest revision as of 01:41, 11 September 2024
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
Solution 5
Let =
By Newton's Sums,
Solving for , we get
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.