2003 AMC 10A Problems/Problem 18

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 $x^2$,

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

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

See Also

2003 AMC 10A (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS