Difference between revisions of "2019 AMC 12A Problems/Problem 17"
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==Solution 1== | ==Solution 1== | ||
− | Applying | + | Applying [https://artofproblemsolving.com/wiki/index.php/Newton's_Sums Newton's Sums], we have<cmath>s_{k+1}+(-5)s_k+(8)s_{k-1}+(-13)s_{k-2}=0,</cmath>so<cmath>s_{k+1}=5s_k-8s_{k-1}+13s_{k-2},</cmath>we get the answer as <math>5+(-8)+13=10</math>. |
==Solution 2== | ==Solution 2== | ||
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==Solution 3== | ==Solution 3== | ||
− | Let <math>p, q</math>, and <math>r</math> be the roots of the polynomial. By | + | Let <math>p, q</math>, and <math>r</math> be the roots of the polynomial. By Vieta's Formulae, we have |
<math>p+q+r = 5</math> | <math>p+q+r = 5</math> | ||
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-IzhanAli | -IzhanAli | ||
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+ | ==Solution 5 == | ||
+ | Let the roots be <math>r</math>, <math>s</math>, and <math>t</math>. We know <math>r^2+s^2+t^2 = (r+s+t)(r+s+t) - 2(rs+st+tr)</math>.Continuing, we have: | ||
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+ | <math>r^3+s^3+t^3 = (r^2+s^2+t^2)(r+s+t) - (rs+st+tr)(r+s+t)+3rst</math> | ||
+ | |||
+ | |||
+ | <math>r^4+s^4+t^4 = (r^3+s^3+t^3)(r+s+t) - (rs+st+tr)(r^2+s^2+t^2)-(rst)(r+s+t)</math> | ||
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+ | <math>r^5+s^5+t^5 = (r^4+s^4+t^4)(r+s+t) - (rs+st+tr)(r^3+s^3+t^3) - (rst)(r^2+s^2+t^2)</math> | ||
+ | |||
+ | |||
+ | Clearly, the answer is <math>5-8+13 = \boxed{\textbf{(D)} 10}</math> | ||
+ | |||
+ | -skibbysiggy | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 12:52, 3 October 2024
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, 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 Vieta'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
-IzhanAli
Solution 5
Let the roots be , , and . We know .Continuing, we have:
Clearly, the answer is
-skibbysiggy
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.