Difference between revisions of "2021 AMC 12B Problems/Problem 16"
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Let the three roots of <math>f(x)</math> be <math>d</math>, <math>e</math>, and <math>f</math>. (Here e does NOT mean 2.7182818...) | Let the three roots of <math>f(x)</math> be <math>d</math>, <math>e</math>, and <math>f</math>. (Here e does NOT mean 2.7182818...) | ||
We know that <math>a=-(d+e+f)</math>, <math>b=de+ef+df</math>, and <math>c=-def</math>, and that <math>g(1)=1-\frac{1}{d}-\frac{1}{e}-\frac{1}{f}+\frac{1}{de}+\frac{1}{ef}+\frac{1}{df}-\frac{1}{def}</math> (Vieta's). This is equal to <math>\frac{def-de-df-ef+d+e+f-1}{def}</math>, which equals <math>\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}</math>. -dstanz5 | We know that <math>a=-(d+e+f)</math>, <math>b=de+ef+df</math>, and <math>c=-def</math>, and that <math>g(1)=1-\frac{1}{d}-\frac{1}{e}-\frac{1}{f}+\frac{1}{de}+\frac{1}{ef}+\frac{1}{df}-\frac{1}{def}</math> (Vieta's). This is equal to <math>\frac{def-de-df-ef+d+e+f-1}{def}</math>, which equals <math>\boxed{(\textbf{A}) \frac{1+a+b+c}{c}}</math>. -dstanz5 | ||
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+ | ==Solution 3 (Fakesolve) == | ||
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+ | Because the problem doesn't specify what the coefficients of the polynomial actually are, we can just plug in any arbitrary polynomial that satisfies the constraints. Let's take <math>f(x) = (x+5)^3 = x^3+15x^2+75x+125</math>. Then <math>f(x)</math> has a triple root of <math>x = -5</math>. Then <math>g(x)</math> has a triple root of <math>-\frac{1}{5}</math>, and it's monic, so <math>g(x) = \left(x+\frac{1}{5}\right)^3 = \frac{125x^3+75x^2+15x+1}{125}</math>. We can see that this is <math>\frac{1+a+b+c}{c}</math>, which is answer choice <math>\boxed{(A)}</math>. | ||
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+ | -Darren Yao | ||
== Video Solution by OmegaLearn (Vieta's Formula) == | == Video Solution by OmegaLearn (Vieta's Formula) == |
Revision as of 15:28, 12 February 2021
Contents
[hide]Problem
Let be a polynomial with leading coefficient whose three roots are the reciprocals of the three roots of where What is in terms of and
Solution 1
Note that has the same roots as , if it is multiplied by some monomial so that the leading term is they will be equal. We have so we can see that Therefore
Solution 2 (Vieta's bash)
Let the three roots of be , , and . (Here e does NOT mean 2.7182818...) We know that , , and , and that (Vieta's). This is equal to , which equals . -dstanz5
Solution 3 (Fakesolve)
Because the problem doesn't specify what the coefficients of the polynomial actually are, we can just plug in any arbitrary polynomial that satisfies the constraints. Let's take . Then has a triple root of . Then has a triple root of , and it's monic, so . We can see that this is , which is answer choice .
-Darren Yao
Video Solution by OmegaLearn (Vieta's Formula)
~ pi_is_3.14
Video Solution by Hawk Math
https://www.youtube.com/watch?v=p4iCAZRUESs
See Also
2021 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 |
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