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 1: / Line 1: @@
 == Problem ==
-Let <math> P </math> be the [[product]] of the [[nonreal]] [[root]]s of <math> x^4-4x^3+6x^2-4x=2005. </math> Find <math> \lfloor P\rfloor. </math>
+Let <math> P </math> be the product of the nonreal roots of <math> x^4-4x^3+6x^2-4x=2005. </math> Find <math> \lfloor P\rfloor. </math>
 == Solution 1 ==
@@ Line 16: / Line 16: @@
 <math>((x-1)^2+\sqrt{2006})((x-1)^2 -\sqrt{2006})=0</math>
 If you think of each part of the product as a quadratic, then <math>((x-1)^2+\sqrt{2006})</math> is bound to hold the two  non-real roots since the other definitely crosses the x-axis twice since it is just <math>x^2</math> translated down and right.
-Therefore the products of the roots of <math>((x-1)^2+\sqrt{2006})</math> or <math> P=1+\sqrt{2006}</math> so
+Therefore <math>P</math> is the product 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 = \boxed{045}</math>.
@@ Line 36: / Line 36: @@
-==Solution 6(De Moivre's Theorem)==
-As all the other solutions, we find that <math>(x-1)^4 = 2006</math>. Thus <math>x=\sqrt[4]{2006}+1</math>. Thus <math>x= \sqrt[4]{2006}(cos(\frac{2\pi(k)}{4}+isin(\frac{2\pi(k)}{4})</math>
+Note: <math>\frac{2005}{\sqrt[4]{2006}-1}=\frac{2005(1+\sqrt[4]{2006})}{2005} = 1+\sqrt[4]{2006}.</math>
+==Solution 6 (De Moivre's Theorem)==
+As all the other solutions, we find that <math>(x-1)^4 = 2006</math>. Thus <math>x=\sqrt[4]{2006}+1</math>. Thus <math>x= \sqrt[4]{2006}(\cos(\frac{2\pi(k)}{4}+i\sin(\frac{2\pi(k)}{4}))+1</math> when <math>k=0,1,2,3</math>. The complex values of <math>x</math> are the ones where <math>i\sin(\frac{2\pi(k)}{4})</math> does not equal 0. These complex roots are <math>1+\sqrt[4]{2006}(i)</math> and <math>1-\sqrt[4]{2006}(i)</math>. The product of these two nonreal roots is (<math>1+\sqrt[4]{2006}(i)</math>)(<math>1-\sqrt[4]{2006}(i)</math>) which is equal to <math>1+\sqrt {2006}</math>. The floor of that value is <math>\boxed{045}</math>.
 == See also ==
+Video Solution
+https://www.youtube.com/watch?v=LbHg1Su2Rmg
 {{AIME box|year=2005|n=I|num-b=5|num-a=7}}
 [[Category:Intermediate Algebra Problems]]
 {{MAA Notice}}

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