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Difference between revisions of "2005 AIME I Problems/Problem 6"

 
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== Problem ==
 
== Problem ==
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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>
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Let <math> P </math> be the [[product]] of the [[nonreal]] [[root]]s of <math> x^4-4x^3+6x^2-4x=2005. </math> Find <math> \lfloor P\rfloor. </math>
  
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== Solution ==
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The left-hand side of that [[equation]] is nearly equal to <math>(x - 1)^4</math>.  Thus, we add 1 to each side in order to complete the fourth power and get
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<math>(x - 1)^4 = 2006</math>.
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Let <math>r = \sqrt[4]{2006}</math> be the positive [[real]] fourth root of 2006.  Then the roots of the above equation are <math>x = 1 + i^n r</math> for <math>n = 0, 1, 2, 3</math>.  The two non-real members of this set are <math>1 + ir</math> and <math>1 - ir</math>.  Their product is <math>\displaystyle P = 1 + r^2 = 1 + \sqrt{2006}</math>.  <math>44^2 = 1936 < 2006 < 2025 = 45^2</math> so <math>\lfloor P \rfloor = 1 + 44 = 045</math>.
  
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== Solution ==
 
  
 
== See also ==
 
== See also ==
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* [[2005 AIME I Problems/Problem 5 | Previous problem]]
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* [[2005 AIME I Problems/Problem 7 | Next problem]]
 
* [[2005 AIME I Problems]]
 
* [[2005 AIME I Problems]]
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[[Category:Intermediate Algebra Problems]]
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[[Category:Intermediate Complex Number Problems]]

Revision as of 13:21, 17 January 2007

Problem

Let $P$ be the product of the nonreal roots of $x^4-4x^3+6x^2-4x=2005.$ Find $\lfloor P\rfloor.$

Solution

The left-hand side of that equation is nearly equal to $(x - 1)^4$. Thus, we add 1 to each side in order to complete the fourth power and get $(x - 1)^4 = 2006$.

Let $r = \sqrt[4]{2006}$ be the positive real fourth root of 2006. Then the roots of the above equation are $x = 1 + i^n r$ for $n = 0, 1, 2, 3$. The two non-real members of this set are $1 + ir$ and $1 - ir$. Their product is $\displaystyle P = 1 + r^2 = 1 + \sqrt{2006}$. $44^2 = 1936 < 2006 < 2025 = 45^2$ so $\lfloor P \rfloor = 1 + 44 = 045$.


See also

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