2005 AIME I Problems/Problem 6
Contents
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 is the product 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 4
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 .
Solution 5
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 .
Note:
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 .
See also
Video Solution https://www.youtube.com/watch?v=LbHg1Su2Rmg
2005 AIME I (Problems • Answer Key • Resources) | ||
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