2005 AIME I Problems/Problem 6
Problem
Let be the product of the nonreal roots of Find
Solution 1
The left-hand side of that equation is nearly equal to . Thus, we add 1 to each side in order to complete the fourth power and get .
Let be the positive real fourth root of 2006. Then the roots of the above equation are for . The two non-real members of this set are and . Their product is . so .
Solution 2
Starting like before, This time we apply differences of squares. so If you think of each part of the product as a quadratic, then is bound to hold the two non-real roots since the other definitely crosses the x-axis twice since it is just translated down and right. Therefore the products of the roots of or so
.
Solution 3
If we don't see the fourth power, we can always factor the LHS to try to create a quadratic substitution. Checking, we find that and are both roots. Synthetic division gives . We now have our quadratic substitution of , giving us . From here we proceed as in Solution 1 to get .
-Solution by thecmd999
See also
2005 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.