Difference between revisions of "2005 AIME I Problems/Problem 6"
(→Solution 6(De Moivre's Theorem)) |
(→Solution 6(De Moivre's Theorem)) |
||
Line 38: | Line 38: | ||
==Solution 6(De Moivre's Theorem)== | ==Solution 6(De Moivre's Theorem)== | ||
− | 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}))+1</math> when <math>k=0,1,2,3</math>. The complex values of x are the ones where <math>isin(\frac{2\pi(k)}{4}</math> does not equal 0. These complex roots are <math>1+\sqrt[4]{2006}(i)</math> and <math>1-\sqrt[4]{2006}(i)</math>. The product of these two nonreal roots is (<math>1+\sqrt[4]{2006}(i)</math>)(<math>1-\sqrt[4]{2006}(i)</math>) which is equal to <math>1+\sqrt {2006}</math>. The floor of that value is | + | 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}))+1</math> when <math>k=0,1,2,3</math>. The complex values of x are the ones where <math>isin(\frac{2\pi(k)}{4}</math> does not equal 0. These complex roots are <math>1+\sqrt[4]{2006}(i)</math> and <math>1-\sqrt[4]{2006}(i)</math>. The product of these two nonreal roots is (<math>1+\sqrt[4]{2006}(i)</math>)(<math>1-\sqrt[4]{2006}(i)</math>) which is equal to <math>1+\sqrt {2006}</math>. The floor of that value is <math>\boxed{045}</math>. |
== See also == | == See also == |
Revision as of 16:45, 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
when
. The complex values of x 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
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.