Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2003 AMC 10A Problems/Problem 18"

(→‎Solution 1)
m
 
(2 intermediate revisions by 2 users not shown)
Line 4: Line 4:
  
 
<math> \mathrm{(A) \ } -\frac{2004}{2003}\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } \frac{2003}{2004}\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } \frac{2004}{2003} </math>
 
<math> \mathrm{(A) \ } -\frac{2004}{2003}\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } \frac{2003}{2004}\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } \frac{2004}{2003} </math>
−
 
−
== Video Solution ==
 
−
https://youtu.be/3dfbWzOfJAI?t=456
 
−
 
−
~ pi_is_3.14
 
  
 
== Solution 1 ==
 
== Solution 1 ==
Line 28: Line 23:
  
 
== Solution 2 ==
 
== Solution 2 ==
−
Dividing both sides by <math>x^2</math>,  
+
Dividing both sides by <math>x</math>,  
  
 
<math>\frac{2003}{2004}+\frac 1x+\frac{1}{x^2}=0</math>,  
 
<math>\frac{2003}{2004}+\frac 1x+\frac{1}{x^2}=0</math>,  
Line 70: Line 65:
  
 
~mathboy282
 
~mathboy282
 +
 +
== Video Solution ==
 +
https://youtu.be/3dfbWzOfJAI?t=456
 +
 +
~ pi_is_3.14
 +
 +
==Video Solution==
 +
https://youtu.be/j1nVw3ASBYU
 +
 +
~savannahsolver
  
 
== See Also ==
 
== See Also ==

Latest revision as of 12:41, 26 August 2023

Problem

What is the sum of the reciprocals of the roots of the equation $\frac{2003}{2004}x+1+\frac{1}{x}=0$?

$\mathrm{(A) \ } -\frac{2004}{2003}\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } \frac{2003}{2004}\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } \frac{2004}{2003}$

Solution 1

Multiplying both sides by $x$:

$\frac{2003}{2004}x^{2}+1x+1=0$

Let the roots be $a$ and $b$.

The problem is asking for $\frac{1}{a}+\frac{1}{b}= \frac{a+b}{ab}$

By Vieta's formulas:

$a+b=(-1)^{1}\frac{1}{\frac{2003}{2004}}=-\frac{2004}{2003}$

$ab=(-1)^{2}\frac{1}{\frac{2003}{2004}}=\frac{2004}{2003}$

So the answer is $\frac{a+b}{ab}=\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}=-1 \Rightarrow\boxed{\mathrm{(B)}\ -1}$.

Solution 2

Dividing both sides by $x$,

$\frac{2003}{2004}+\frac 1x+\frac{1}{x^2}=0$,

we see by Vieta's formulas that the sum of the roots is $-1 \Rightarrow\boxed{\mathrm{(B)}\ -1}$.

Solution 3(Alternate Approach to Solution 1)

Re-stating and multiplying $\frac{1}{x}$ by $\frac{2004}{2004}$, we have $\frac{2003x}{2004} + \frac{2004}{2004x} + 1 = 0$.

Simplifying, we have $\frac{2003}{2004}+\frac{\frac{2004}{x}}{2004}=-1.$

Putting this together, we have $\frac{2003+\frac{2004}{x}}{2004}=-1.$

Simplifying, we have $2004x + \frac{2004}{x} = -2004.$

Multiplying both sides by $x,$ we have $2003x^2+2004=-2004x.$

Moving $-2004x$ to the left side, we have $2003x^2 + 2004x + 2004 = 0.$

Let's go back and see what we want.

We want the sum of the reciprocals of the roots, in which if the roots were to be called $r_1$ and $r_2,$ we would want $\frac{1}{r_1}+\frac{1}{r_2}=\frac{r_1 + r_2}{r_1 \cdot r_2}$.

We can use Vieta's formulas to solve the sum of the roots and the product of the roots.

We have that the sum of the two roots is $\frac{-b}{a}$ where the quadratic is $ax^2+bx+c=0.$

In this case, $a=2003$ and $b=2004$.

Therefore, $r_1+r_2=-\frac{2004}{2003}.$

We have that the product of the two roots is $\frac{c}{a}$ where the quadratic is $ax^2+bx+c=0.$

In this case, $c=2004$ and $a=2003.$

Therefore, $r_1 \cdot r_2 = \frac{2004}{2003}.$

Now that we have the sum and product of the roots, we can substitute back into the expression of what we want.

We have $\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}=-\frac{2004}{2003} \cdot \frac{2003}{2004} = \boxed{\mathrm{(B)}\ -1}.$

~mathboy282

Video Solution

https://youtu.be/3dfbWzOfJAI?t=456

~ pi_is_3.14

Video Solution

https://youtu.be/j1nVw3ASBYU

~savannahsolver

See Also

2003 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2003_AMC_10A_Problems/Problem_18&oldid=197500"