Difference between revisions of "2003 AIME II Problems/Problem 9"
Jonathan.li (talk | contribs) (→Solution 5) |
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== Solution 5 == | == Solution 5 == | ||
− | <math> | + | Let <math>S_k</math> = <math>z_1^k+z_2^k+z_3^k+z_4^k</math> |
− | + | By [[Newton's Sums]], | |
− | |||
− | By Newton's Sums, | ||
<math>S_1-1=0</math> | <math>S_1-1=0</math> | ||
Line 68: | Line 66: | ||
<math>S_4-S_3-S_2-4=0</math> | <math>S_4-S_3-S_2-4=0</math> | ||
− | <math>S_5-S_4-S_3- | + | <math>S_5-S_4-S_3-S_1=0</math> |
+ | |||
+ | <math>S_6-S_5-S_4-S_2=0</math> | ||
− | <math>S_6 | + | 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 == | == Video Solution by Sal Khan == |
Latest revision as of 02:41, 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 |
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