AMC 10 · 2003 · #18

Grade 8 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Symmetric Function of Roots via Vieta

vieta-formulasquadratic-equationsfraction-arithmetic convert-to-algebra ↑ Prerequisites: quadratic-equations

📏 Medium solution 💡 2 insights

Problem

The equation

\frac{2003}{2004}x+1+\frac{1}{x}=0

has two roots, call them

r

and

s

. Find the sum of their reciprocals,

\frac{1}{r}+\frac{1}{s}

Pick an answer.

(A)

$-\frac{2004}{2003}$

(B)

-1

(C)

$\frac{2003}{2004}$

(D)

(E)

$\frac{2004}{2003}$

AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

The equation mixes an x term with a 1/x term, so Tool #13 (Convert to Algebra) first clears the fraction and exposes a plain quadratic. Tool #4 (Introduce a Variable) names the two roots r and s so we can talk about them without solving for them. The decisive move is Tool #16 (Change Focus): instead of hunting for the roots and then taking reciprocals, we notice that the sum of reciprocals 1/r+1/s depends only on the sum r+s and the product rs — two numbers a quadratic hands over directly through its coefficients.

1STEP 1

Clear the fraction to get a quadratic

Multiply every term by $x$ (no root is lost, since $x=0$ never was one), then by 2004: $2003x^{2}+2004x+2004=0$ .

2003/2004x² + x + 1 = 0 → 2003x² + 2004x + 2004 = 0

2STEP 2

Rewrite the target with sum and product

Over a common denominator, $\frac{1}{r}+\frac{1}{s}=\frac{r+s}{rs}$ : only the sum and the product are needed.

1/r+1/s=(r+s)/rs

3STEP 3

Read off the sum and the product

Matching $a(x-r)(x-s)$ with the cleared quadratic gives $r+s=-\frac{2004}{2003}$ and $rs=\frac{2004}{2003}$ .

r+s=-2004/2003, rs=2004/2003

4STEP 4

Divide to get the answer

Top and bottom are the same size with opposite signs, so $\frac{r+s}{rs}=-1$ — choice (B).

(r+s)/rs=-2004/2003/2004/2003=-1 → (B)

Answer

-1

The clean cancellation is a sign the problem was built this way on purpose. For any quadratic ax²+bx+c=0, the sum of the reciprocals of the roots is (r+s)/rs=(-b/a)/(c/a)=-b/c — the leading coefficient a never enters. Here b=2004 and c=2004, so the value is -2004/2004=-1 no matter what the leading 2003 is. The awkward fraction 2003/2004 was a distraction; the equal middle and constant coefficients pin the answer to -1. Relating coefficients to the sum and product of roots is Algebra I material, but the arithmetic itself is middle-school fraction work.

💡Key takeaway

The sum of the reciprocals of a quadratic's roots is just -b/c (the middle coefficient over the constant), so you can answer without ever finding the roots.

Clear the fraction to get a quadratic
Rewrite the target with sum and product
Read off the sum and the product
Divide to get the answer

Clear the fraction to get a quadratic

Rewrite the target with sum and product

Read off the sum and the product

Divide to get the answer

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