Difference between revisions of "2005 AIME I Problems/Problem 6"
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== Problem == | == Problem == | ||
− | Let <math> P </math> be the | + | Let <math> P </math> be the product of the nonreal roots of <math> x^4-4x^3+6x^2-4x=2005. </math> Find <math> \lfloor P\rfloor. </math> |
== Solution 1 == | == Solution 1 == | ||
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<math>((x-1)^2+\sqrt{2006})((x-1)^2 -\sqrt{2006})=0</math> | <math>((x-1)^2+\sqrt{2006})((x-1)^2 -\sqrt{2006})=0</math> | ||
If you think of each part of the product as a quadratic, then <math>((x-1)^2+\sqrt{2006})</math> is bound to hold the two non-real roots since the other definitely crosses the x-axis twice since it is just <math>x^2</math> translated down and right. | If you think of each part of the product as a quadratic, then <math>((x-1)^2+\sqrt{2006})</math> is bound to hold the two non-real roots since the other definitely crosses the x-axis twice since it is just <math>x^2</math> translated down and right. | ||
− | Therefore the | + | Therefore <math>P</math> is the product of the roots of <math>((x-1)^2+\sqrt{2006})</math> or <math> P=1+\sqrt{2006}</math> so |
<math>\lfloor P \rfloor = 1 + 44 = \boxed{045}</math>. | <math>\lfloor P \rfloor = 1 + 44 = \boxed{045}</math>. | ||
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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}+ | + | Note: <math>\frac{2005}{\sqrt[4]{2006}-1}=\frac{2005(1+\sqrt[4]{2006})}{2005} = 1+\sqrt[4]{2006}.</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}+i\sin(\frac{2\pi(k)}{4}))+1</math> when <math>k=0,1,2,3</math>. The complex values of <math>x</math> are the ones where <math>i\sin(\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>. | ||
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== See also == | == See also == | ||
+ | Video Solution | ||
+ | https://www.youtube.com/watch?v=LbHg1Su2Rmg | ||
{{AIME box|year=2005|n=I|num-b=5|num-a=7}} | {{AIME box|year=2005|n=I|num-b=5|num-a=7}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 12:52, 13 June 2024
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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Followed by Problem 7 | |
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