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

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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>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>.
 
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>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>.
 +
 +
Alternate Solution:
 +
Starting like before
 +
<math>(x-1)^4= 2006</math>
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This time we apply differences of squares.
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<math>(x-1)^4-2006=0</math> so
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<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.
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Therefore the products of the roots of <math>((x-1)^2+\sqrt{2006})</math> or <math> P=1+\sqrt{2006}</math> so
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<math>\lfloor P \rfloor = 1 + 44 = 045</math>.
  
 
== See also ==
 
== See also ==

Revision as of 16:02, 26 November 2011

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 $P = 1 + r^2 = 1 + \sqrt{2006}$. $44^2 = 1936 < 2006 < 2025 = 45^2$ so $\lfloor P \rfloor = 1 + 44 = 045$.

Alternate Solution: Starting like before $(x-1)^4= 2006$ This time we apply differences of squares. $(x-1)^4-2006=0$ so $((x-1)^2+\sqrt{2006})((x-1)^2 -\sqrt{2006})=0$ If you think of each part of the product as a quadratic, then $((x-1)^2+\sqrt{2006})$ is bound to hold the two non-real roots since the other definitely crosses the x-axis twice since it is just $x^2$ translated down and right. Therefore the products of the roots of $((x-1)^2+\sqrt{2006})$ or $P=1+\sqrt{2006}$ so

$\lfloor P \rfloor = 1 + 44 = 045$.

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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