2005 AIME I Problems/Problem 6
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 .
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
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 .
Realizing that if we add 1 to both sides we get which can be factored as . Then we can substitute with which leaves us with . Now subtracting 2006 from both sides we get some difference of squares . The question asks for the product of the complex roots so we only care about the last factor which is equal to zero. From there we can solve , we can substitute for giving us , expanding this we get . We know that the product of a quadratics roots is which leaves us with .
As in solution 1, we find that . Now so and are the real roots of the equation. Multiplying, we get . Now transforming the original function and using Vieta's formula, so . We find that the product of the nonreal roots is and we get .
Solution 6 (De Moivre's Theorem)
As all the other solutions, we find that . Thus . Thus when . The complex values of are the ones where does not equal 0. These complex roots are and . The product of these two nonreal roots is ()() which is equal to . The floor of that value is .
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