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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>.
  
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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>.
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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 = \boxed{045}</math>.
  
 
== Solution 2 ==
 
== Solution 2 ==

Revision as of 14:15, 3 November 2015

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

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 3

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 $\boxed{045}$.

-Solution by thecmd999

Solution 4

Realizing that if we add 1 to both sides we get $x^4-4x^3+6x^2-4x+1=2006$ which can be factored as $(x-1)^4=2006$. Then we can substitute $(x-1)$ with $y$ which leaves us with $y^4=2006$. Now subtracting 2006 from both sides we get some difference of squares $y^4-2006=0 \rightarrow (y-\sqrt[4]{2006})(y+\sqrt[4]{2006})(y^2+\sqrt{2006})=0$. The question asks for the product of the complex roots so we only care about the last factor which is equal to zero. From there we can solve $y^2+\sqrt{2006}=0$, we can substitute $(x-1)$ for $y$ giving us $(x-1)^2+\sqrt{2006}=0$, expanding this we get $x^2-2x+1+\sqrt{2006}=0$. We know that the product of a quadratics roots is $\frac{c}{a}$ which leaves us with $\frac{1+\sqrt{2006}}{1}=1+\sqrt{2006}\approx\boxed{045}$.

Solution 5

As in solution 1, we find that $(x-1)^4 = 2006$. Now $x-1=\pm \sqrt[4]{2006}$ so $x_1 = 1+\sqrt[4]{2006}$ and $x_2 = 1-\sqrt[4]{2006}$ are the real roots of the equation. Multiplying, we get $x_1 x_2 = 1 - \sqrt{2006}$. Now transforming the original function and using Vieta's formula, $x^4-4x^3+6x^2-4x-2005=0$ so $x_1 x_2 x_3 x_4 = \frac{-2005}{1} = -2005$. We find that the product of the nonreal roots is $x_3 x_4 = \frac{-2005}{1-\sqrt{2006}} \approx 45.8$ and we get $\boxed{045}$.

See also

2005 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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All AIME Problems and Solutions

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