Difference between revisions of "2014 AIME I Problems/Problem 9"

(Solution)
Line 7: Line 7:
 
Let <math>x_1< x_2 < x_3</math> be the three real roots of the equation <math>\sqrt{2014} x^3 - 4029x^2 + 2 = 0</math>. Find <math>x_2(x_1+x_3)</math>.
 
Let <math>x_1< x_2 < x_3</math> be the three real roots of the equation <math>\sqrt{2014} x^3 - 4029x^2 + 2 = 0</math>. Find <math>x_2(x_1+x_3)</math>.
  
We note that <math>\dfrac{1}{\sqrt{2014}}</math> is a solution since <math>(\dfrac{1}{\sqrt{2014}})^3*\sqrt{2014}-4029(\dfrac{1}{\sqrt{2014}})^2+2=\dfrac{1}{2014}-\dfrac{4029}{2014}+2=\dfrac{1-4029+4028}{2014} = 0</math>
+
Substituting <math>n</math> for <math>2014</math>, we get <math>\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2 = x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0</math>. Noting that <math>nx^2 - 1</math> factors as a difference of squares to <math>(\sqrt{n}x - 1)(\sqrt{n}x+1)</math>, we can factor the left side as <math>(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))</math>. This means that <math>\frac{1}{\sqrt{n}}</math> is a root, and the other two roots are the roots of <math>x^2 - 2\sqrt{n}x - 2</math>. Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to <math>2\sqrt{n}</math>, so the positive root must be greater than <math>2\sqrt{n}</math> in order to produce this sum when added to a negative value. Since <math>0 < \frac{1}{\sqrt{2014}} < 2\sqrt{2014}</math> is clearly true, <math>x_2 = \frac{1}{\sqrt{2014}}</math> and <math>x_1 + x_3 = 2\sqrt{2014}</math>. Multiplying these values together, we find that <math>x_2(x_1+x_3) = \boxed{002}</math>.
 
 
We claim that <math>x_2=\dfrac{1}{\sqrt{2014}}</math>
 
[please clarify]
 
 
 
By Vieta's formula we have that the <math>x^2</math> coefficent is equal to <math>-x_1-x_2-x_3</math> and that the <math>x</math> coefficent is equal to <math>x_1x_2+x_1x_3+x_2x_3</math>. Using the values in the above equation we get:
 
<math>-x_1-x_2-x_3=-4029\rightarrow x_1+x_2+x_3=4029</math>
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2014|n=I|num-b=8|num-a=10}}
 
{{AIME box|year=2014|n=I|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:12, 14 March 2014

Problem 9

Let $x_1<x_2<x_3$ be the three real roots of the equation $\sqrt{2014}x^3-4029x^2+2=0$. Find $x_2(x_1+x_3)$.

Solution

Let $x_1< x_2 < x_3$ be the three real roots of the equation $\sqrt{2014} x^3 - 4029x^2 + 2 = 0$. Find $x_2(x_1+x_3)$.

Substituting $n$ for $2014$, we get $\sqrt{n}x^3 - (1+2n)x^2 + 2 = \sqrt{n}x^3 - x^2 - 2nx^2 + 2 = x^2(\sqrt{n}x - 1) - 2(nx^2 - 1) = 0$. Noting that $nx^2 - 1$ factors as a difference of squares to $(\sqrt{n}x - 1)(\sqrt{n}x+1)$, we can factor the left side as $(\sqrt{n}x - 1)(x^2 - 2(\sqrt{n}x+1))$. This means that $\frac{1}{\sqrt{n}}$ is a root, and the other two roots are the roots of $x^2 - 2\sqrt{n}x - 2$. Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to $2\sqrt{n}$, so the positive root must be greater than $2\sqrt{n}$ in order to produce this sum when added to a negative value. Since $0 < \frac{1}{\sqrt{2014}} < 2\sqrt{2014}$ is clearly true, $x_2 = \frac{1}{\sqrt{2014}}$ and $x_1 + x_3 = 2\sqrt{2014}$. Multiplying these values together, we find that $x_2(x_1+x_3) = \boxed{002}$.

See also

2014 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 8
Followed by
Problem 10
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS