Difference between revisions of "2019 AMC 10A Problems/Problem 24"
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Multiplying both sides by <math>(s-p)(s-q)(s-r)</math> yields | Multiplying both sides by <math>(s-p)(s-q)(s-r)</math> yields | ||
<cmath>1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)</cmath> | <cmath>1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)</cmath> |
Revision as of 19:58, 22 November 2020
Problem
Let , , and be the distinct roots of the polynomial . It is given that there exist real numbers , , and such that for all . What is ?
Solution
Multiplying both sides by yields As this is a polynomial identity, and it is true for infinitely many , it must be true for all (since a polynomial with infinitely many roots must in fact be the constant polynomial ). This means we can plug in to find that . Similarly, we can find and . Summing them up, we get that By Vieta's Formulas, we know that and . Thus the answer is .
Note: this process of substituting in the 'forbidden' values in the original identity is a standard technique for partial fraction decomposition, as taught in calculus classes.
See Also
Video Solution: https://www.youtube.com/watch?v=GI5d2ZN8gXY&t=53s
2019 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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All AMC 10 Problems and Solutions |
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