Difference between revisions of "2021 AMC 12B Problems/Problem 16"
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<math>\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}</math> | <math>\textbf{(A) }\frac{1+a+b+c}c \qquad \textbf{(B) }1+a+b+c \qquad \textbf{(C) }\frac{1+a+b+c}{c^2}\qquad \textbf{(D) }\frac{a+b+c}{c^2} \qquad \textbf{(E) }\frac{1+a+b+c}{a+b+c}</math> | ||
− | ==Solution== | + | ==Solution 1== |
Note that <math>f(1/x)</math> has the same roots as <math>g(x)</math>, if it is multiplied by some monomial so that the leading term is <math>x^3</math> they will be equal. We have | Note that <math>f(1/x)</math> has the same roots as <math>g(x)</math>, if it is multiplied by some monomial so that the leading term is <math>x^3</math> they will be equal. We have | ||
<cmath>f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c</cmath> | <cmath>f(1/x) = \frac{1}{x^3} + \frac{a}{x^2}+\frac{b}{x} + c</cmath> |
Revision as of 12:11, 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
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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