2005 AIME I Problems/Problem 6
Problem
Let be the product of the nonreal roots of Find
Solution
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 .
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 |