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Difference between revisions of "2003 AMC 10A Problems/Problem 18"
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<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>
  
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== Solution ==
+
== Solution 1 ==
 
Multiplying both sides by <math>x</math>:  
 
Multiplying both sides by <math>x</math>:  
  
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<math>ab=(-1)^{2}\frac{1}{\frac{2003}{2004}}=\frac{2004}{2003}</math>
 
<math>ab=(-1)^{2}\frac{1}{\frac{2003}{2004}}=\frac{2004}{2003}</math>
  
−
So the answer is <math> \frac{a+b}{ab}=\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}=-1 \Rightarrow B</math>.   
+
So the answer is <math> \frac{a+b}{ab}=\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}=-1 \Rightarrow\boxed{\mathrm{(B)}\ -1}</math>.
 +
 
 +
== Solution 2 ==
 +
Dividing both sides by <math>x</math>,
 +
 
 +
<math>\frac{2003}{2004}+\frac 1x+\frac{1}{x^2}=0</math>,
 +
 
 +
we see by [[Vieta's formulas]] that the sum of the roots is <math>-1 \Rightarrow\boxed{\mathrm{(B)}\ -1}</math>.
 +
 
 +
== Solution 3(Alternate Approach to Solution 1) ==
 +
Re-stating and multiplying <math>\frac{1}{x}</math> by <math>\frac{2004}{2004}</math>, we have <math>\frac{2003x}{2004} + \frac{2004}{2004x} + 1 = 0</math>.
 +
 
 +
Simplifying, we have <math>\frac{2003}{2004}+\frac{\frac{2004}{x}}{2004}=-1.</math>
 +
 
 +
Putting this together, we have <math>\frac{2003+\frac{2004}{x}}{2004}=-1.</math>
 +
 
 +
Simplifying, we have <math>2004x + \frac{2004}{x} = -2004.</math>
 +
 
 +
Multiplying both sides by <math>x,</math> we have <math>2003x^2+2004=-2004x.</math>
 +
 
 +
Moving <math>-2004x</math> to the left side, we have <math>2003x^2 + 2004x + 2004 = 0.</math>
 +
 
 +
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 <math>r_1</math> and <math>r_2,</math> we would want <math>\frac{1}{r_1}+\frac{1}{r_2}=\frac{r_1 + r_2}{r_1 \cdot r_2}</math>.
 +
 
 +
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 <math>\frac{-b}{a}</math> where the quadratic is <math>ax^2+bx+c=0.</math>
 +
 
 +
In this case, <math>a=2003</math> and <math>b=2004</math>.
 +
 
 +
Therefore, <math>r_1+r_2=-\frac{2004}{2003}.</math>
 +
 
 +
We have that the product of the two roots is <math>\frac{c}{a}</math> where the quadratic is <math>ax^2+bx+c=0.</math>
 +
 
 +
In this case, <math>c=2004</math> and <math>a=2003.</math>
 +
 
 +
Therefore, <math>r_1 \cdot r_2 = \frac{2004}{2003}.</math>
 +
 
 +
Now that we have the sum and product of the roots, we can substitute back into the expression of what we want.
 +
 
 +
We have <math>\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}=-\frac{2004}{2003} \cdot \frac{2003}{2004} = \boxed{\mathrm{(B)}\ -1}.</math>
 +
 
 +
~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
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All AMC 10 Problems and Solutions

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