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

Latest revision as of 11:52, 13 June 2024

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 $P$ is the product 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 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 $\boxed{045}$ .

Note: $\frac{2005}{\sqrt[4]{2006}-1}=\frac{2005(1+\sqrt[4]{2006})}{2005} = 1+\sqrt[4]{2006}.$

Solution 6 (De Moivre's Theorem)

As all the other solutions, we find that $(x-1)^4 = 2006$ . Thus $x=\sqrt[4]{2006}+1$ . Thus $x= \sqrt[4]{2006}(\cos(\frac{2\pi(k)}{4}+i\sin(\frac{2\pi(k)}{4}))+1$ when $k=0,1,2,3$ . The complex values of $x$ are the ones where $i\sin(\frac{2\pi(k)}{4})$ does not equal 0. These complex roots are $1+\sqrt[4]{2006}(i)$ and $1-\sqrt[4]{2006}(i)$ . The product of these two nonreal roots is ( $1+\sqrt[4]{2006}(i)$ )( $1-\sqrt[4]{2006}(i)$ ) which is equal to $1+\sqrt {2006}$ . The floor of that value is $\boxed{045}$ .

@@ Line 44: / Line 44: @@
-== Solution 7 ==
-First, set one side to 0: <math>x^4-4x^3+6x^2-4x-2005=0</math>. We can try to locate the range for the real roots by setting values for <math>x</math>. When <math>x=7</math>, <math>f(x)=-710</math> and when <math>x=8</math>, <math>f(x)=395</math>. From this, we know that there is a real root between <math>7</math> and <math>8</math>.
-We move on to the negative <math>x</math> values. When <math>x=-5</math>, <math>f(x)=-710</math> and when <math>x=-6</math>, <math>f(x)=395</math>. From this, we know that there is a real root between <math>-5</math> and <math>-6</math>.
-Using Vieta's formulas, the product of all the roots is <math>-2005</math>. We can find the upper and the lower bound of <math>P</math> by dividing <math>-2005</math> by the range values. To get the lower bound, we divide <math>-2005</math> by <math>8</math> and <math>-6</math> to get <math>41.77</math>. To get the upper bound, we divide <math>-2005</math> by <math>7</math> and <math>-5</math> to get <math>57.28</math>. We now that <math>P</math> must be between <math>41.77</math> and <math>57.28</math>. There are too many values in this range, so we further try to narrow it down.
-Going back to the equation, we guess <math>x=7.7</math>, so <math>f(x)=9.11</math>. Since this is close to <math>0</math>, we guess <math>x=7.69</math> so <math>f(x)=-2.89</math>. From this, we know that there is a real root between <math>7.7</math> and <math>7.69</math>. This is much more narrow than <math>7</math> and <math>8</math>.
-Using the same process, the negative <math>x</math> values are <math>x=-5.7</math> and <math>x=-5.69</math>, giving <math>f(x)=9.11</math> and <math>f(x)=-2.89</math> respectively.
-We find the new lower bound of <math>P</math> by dividing <math>-2005</math> by <math>7.7</math> and <math>-5.7</math>, and the new upper bound by dividing <math>-2005</math> by <math>7.69</math> and <math>-5.69</math> to get <math>45.6</math> and <math>45.8</math> respectively.
-From this, we know that <math>\lfloor P \rfloor = \boxed{045}</math>.
 == 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

Page

Toolbox

Search