2019 AMC 10A Problems/Problem 24
Contents
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 1
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 We can express , and by Vieta's Formulas, we know that this expression is equal to . Vieta's also gives (which we also used to find ), so 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.
-very small latex edit from countmath1 :)
Minor rephrasing for correctness and clarity ~ Technodoggo
Solution 2 (Pure Elementary Algebra)
Solution 1 uses a trick from Calculus that seemingly contradicts the restriction . I am going to provide a solution with pure elementary algebra. From we get , , , substituting them in , we get
,
,
,
, by symmetry, ,
The rest is similar to solution 1, we get
Video Solution by Richard Rusczyk
https://www.youtube.com/watch?v=GI5d2ZN8gXY
Video Solution by TheBeautyofMath
~IceMatrix
See Also
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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