2014 AIME II Problems/Problem 5
Problem
Real numbers and are roots of , and and are roots of . Find the sum of all possible values of .
Hint
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
The roots of are , , and since they sum to by Vieta's Formula (co-efficient of term is ).
Similarly, the roots of are , , and , as they too sum to .
Then:
and from and
and from .
From these equations, we can write that and simplifying gives
We now move to the other two equations regarding the product of the roots. We see that we can cancel a negative from both sides to get Subtracting the first equation from the second equation gives us .
Expanding, simplifying, substituting , and simplifying some more yields the simple quadratic , so . Then .
Finally, we substitute back into to get , or .
The answer is .
Solution 3
By Vieta's, we know that the sum of roots of is . Therefore, the roots of are . By similar reasoning, the roots of are . Thus, and .
Since and have the same coefficient for , we can go ahead and match those up to get
At this point, we can go ahead and compare the constant term in and . Doing so is certainly valid, but we can actually do this another way. Notice that . Therefore, . If we plug that into our expression, we get that This tells us that or . Since is the product of the roots, we have that the two possibilities are and . Adding the absolute values of these gives us .
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.