Difference between revisions of "2005 AIME I Problems/Problem 6"
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As in solution 1, we find that <math>(x-1)^4 = 2006</math>. Now <math>x-1=\pm \sqrt[4]{2006}</math> so <math>x_1 = 1+\sqrt[4]{2006}</math> and <math>x_2 = 1-\sqrt[4]{2006}</math> are the real roots of the equation. Multiplying, we get <math>x_1 x_2 = 1 - \sqrt{2006}</math>. Now transforming the original function and using Vieta's formula, <math>x^4-4x^3+6x^2-4x-2005=0</math> so <math>x_1 x_2 x_3 x_4 = \frac{-2005}{1} = -2005</math>. We find that the product of the nonreal roots is <math>x_3 x_4 = \frac{-2005}{1-\sqrt{2006}} \approx 45.8</math> and we get <math>\boxed{045}</math>. | As in solution 1, we find that <math>(x-1)^4 = 2006</math>. Now <math>x-1=\pm \sqrt[4]{2006}</math> so <math>x_1 = 1+\sqrt[4]{2006}</math> and <math>x_2 = 1-\sqrt[4]{2006}</math> are the real roots of the equation. Multiplying, we get <math>x_1 x_2 = 1 - \sqrt{2006}</math>. Now transforming the original function and using Vieta's formula, <math>x^4-4x^3+6x^2-4x-2005=0</math> so <math>x_1 x_2 x_3 x_4 = \frac{-2005}{1} = -2005</math>. We find that the product of the nonreal roots is <math>x_3 x_4 = \frac{-2005}{1-\sqrt{2006}} \approx 45.8</math> and we get <math>\boxed{045}</math>. | ||
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+ | ==Solution 6(De Moivre's Theorem)== | ||
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+ | As all the other solutions, we find that <math>(x-1)^4 = 2006</math>. Thus <math>x=\sqrt[4]{2006}+1</math>. Thus <math>x= \sqrt[4]{2006}(cos(\frac{2\pi(k)}{4}+isin(\frac{2\pi(k)}{4})</math> | ||
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== See also == | == See also == | ||
{{AIME box|year=2005|n=I|num-b=5|num-a=7}} | {{AIME box|year=2005|n=I|num-b=5|num-a=7}} |
Revision as of 15:39, 26 July 2018
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 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 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 .
Solution 6(De Moivre's Theorem)
As all the other solutions, we find that . Thus . Thus
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 |
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