Difference between revisions of "2005 AIME I Problems/Problem 6"
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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 | Therefore the products 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 = 045</math>. | + | <math>\lfloor P \rfloor = 1 + 44 = \boxed{045}</math>. |
== See also == | == See also == | ||
Revision as of 21:57, 28 December 2011
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 
.
Alternate Solution:
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
.
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 | ||
