2005 AIME I Problems/Problem 6

Problem

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

Solution 1

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

Solution 2

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

$\lfloor P \rfloor = 1 + 44 = \boxed{045}$ .

Solution 2

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 $x=0$ and $x=2$ are both roots. Synthetic division gives $(x^2-2x)(x^2-2x+2)=2005$ . We now have our quadratic substitution of $y=x^2-2x+1=(x-1)^2$ , giving us $(y-1)(y+1)=2005$ . From here we proceed as in Solution 1 to get $\boxed{045}$ .

2005 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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2005 AIME I Problems/Problem 6

Contents

Problem

Solution 1

Solution 2

Solution 2

See also