2014 AIME II Problems/Problem 5
Contents
Problem
Real numbers and are roots of , and and are roots of . Find the sum of all possible values of .
Solution 1
Let , , and be the roots of (per Vieta's). Then and similarly for . Also,
Set up a similar equation for :
Simplifying and adding the equations gives
Now, let's deal with the terms. Plugging the roots , , and into yields a long polynomial, and plugging the roots , , and into yields another long polynomial. Equating the coefficients of x in both polynomials: which eventually simplifies to
Substitution into (*) should give and , corresponding to and , and , for an answer of .
Solution 2
As above, we know from Vieta's that the roots of are , , and . Similarly, the roots of are , , and . Then and from and and from .
From these equations, we can write that , and simplifying gives us or .
We now move to the other two equations. We see that we can cancel a negative from both sides to get and . Subtracting the first from the second equation gives us . Expanding and simplifying, substituting and simplifying some more yields the simple quadratic , so . Then .
Finally, we substitute back in to get or . Then the answer is .
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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