Difference between revisions of "2005 AIME I Problems/Problem 6"
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<math>(x - 1)^4 = 2006</math>. | <math>(x - 1)^4 = 2006</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> | + | 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>. |
== See also == | == See also == | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
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Revision as of 21:17, 30 November 2007
Problem
Let 
be the product of the nonreal roots of 
Find 
Solution
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 ![$r = \sqrt[4]{2006}$](http://latex.artofproblemsolving.com/6/e/a/6ea78ab025a46a9201473cb8c0fbbaca3b79df50.png)
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 
.
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 | ||
