Difference between revisions of "2019 AMC 12A Problems/Problem 17"
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+ | ==Solution 4== | ||
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+ | Let <math>r,s,t</math> be the roots of <math>x^3-5x^2+8x-13</math>. Then: | ||
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+ | <math>r^3=5r^2-8r+13</math> \ | ||
+ | <math>s^3=5s^2-8s+13</math> \ | ||
+ | <math>t^3=5t^2-8t+13</math> | ||
+ | |||
+ | If we multiply both sides of the equation by <math>r^k</math>, where <math>k</math> is a positive integer, then that won't change the coefficients, but just the degree of the new polynomial and the other term's exponents. We can try multiplying to find <math>r^4+s^4+t^4</math>, but that is just to check. So then with the above information about <math>r^3,s^3,t^3</math>, we see that: | ||
+ | |||
+ | <math>r^k=5r^{k-1}-8r^{k-2}-13r^{k-3}</math>, | ||
+ | <math>s^k=5s^{k-1}-8s^{k-2}-13s^{k-3}</math>, | ||
+ | <math>t^k=5t^{k-1}-8t^{k-2}-13t^{k-3}</math> | ||
+ | |||
+ | <math>s_k=r^k+s^k+t^k</math> | ||
+ | |||
+ | Then: <math>s_k=5s_{k-1}-8s_{k-2}+13s_{k-3}</math> | ||
+ | |||
+ | This means that <math>s_{k+1}=5s_{k}-8s_{k-1}+13s_{k-2}</math>, as expected. So we have <math>a=5, b=-8, c=13</math>. So our answer is <math>5-8+13=\boxed{\textbf{(D) } 10}</math> | ||
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==Video Solution== | ==Video Solution== |
Revision as of 20:59, 13 January 2022
Contents
[hide]Problem
Let denote the sum of the th powers of the roots of the polynomial . In particular, , , and . Let , , and be real numbers such that for , , What is ?
Solution 1
Applying Newton's Sums (see this link), we havesowe get the answer as .
Solution 2
Let , and be the roots of the polynomial. Then,
Adding these three equations, we get
can be written as , giving
We are given that is satisfied for , , , meaning it must be satisfied when , giving us .
Therefore, , and by matching coefficients.
.
Solution 3
Let , and be the roots of the polynomial. By Viète's Formulae, we have
.
We know . Consider .
Using and , we see .
We have
Rearrange to get
So, .
-gregwwl
Solution 4
Let be the roots of . Then:
\ \
If we multiply both sides of the equation by , where is a positive integer, then that won't change the coefficients, but just the degree of the new polynomial and the other term's exponents. We can try multiplying to find , but that is just to check. So then with the above information about , we see that:
, ,
Then:
This means that , as expected. So we have . So our answer is
Video Solution
For those who want a video solution: https://www.youtube.com/watch?v=tAS_DbKmtzI
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.