AMC 10 · 2003 · #18
Grade 8 algebraWhat is the sum of the reciprocals of the roots of the equation
20042003x+1+x1=0?
Pick an answer.
AMC 10 2003 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: The equation $\frac{2003}{2004}x + 1 + \frac{1}{x} = 0$ has two roots. Find the sum of the reciprocals of those two roots.
Givens: The equation is $\frac{2003}{2004}x + 1 + \frac{1}{x} = 0$; It has two roots; call them $r$ and $s$; Answer choices: (A) $-\frac{2004}{2003}$, (B) $-1$, (C) $\frac{2003}{2004}$, (D) $1$, (E) $\frac{2004}{2003}$
Unknowns: The sum of the reciprocals of the roots, $\frac{1}{r}+\frac{1}{s}$
Understand
Restated: The equation $\frac{2003}{2004}x + 1 + \frac{1}{x} = 0$ has two roots. Find the sum of the reciprocals of those two roots.
Givens: The equation is $\frac{2003}{2004}x + 1 + \frac{1}{x} = 0$; It has two roots; call them $r$ and $s$; Answer choices: (A) $-\frac{2004}{2003}$, (B) $-1$, (C) $\frac{2003}{2004}$, (D) $1$, (E) $\frac{2004}{2003}$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #13 Convert to Algebra, #16 Change Focus / Count the Complement, #7 Identify Subproblems
The equation mixes an $x$ term with a $\frac{1}{x}$ 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 $\frac{1}{r}+\frac{1}{s}$ depends only on the sum $r+s$ and the product $rs$ — two numbers a quadratic hands over directly through its coefficients.
Execute — Answer: B
8.EE.C.7 Step 1 Clear the fraction to get a quadratic
- The $\frac{1}{x}$ term hides the fact that this is really a quadratic, so multiply every term by $x$.
- Because $\frac{1}{x}$ appears, $x=0$ is not allowed, so multiplying by $x$ throws away no root.
- This gives $\frac{2003}{2004}x^2 + x + 1 = 0$.
- To clear the remaining fraction, multiply through by $2004$: $2003x^2 + 2004x + 2004 = 0$.
- Now it is an ordinary quadratic with whole-number coefficients.
💡 Multiplying by $x$ turns the $\frac{1}{x}$ term into a plain number and reveals the quadratic hiding underneath.
6.EE.A.3 Step 2 Rewrite the target with sum and product
- Call the two roots $r$ and $s$.
- The quantity we want is $\frac{1}{r}+\frac{1}{s}$.
- Put it over a common denominator: $\frac{1}{r}+\frac{1}{s}=\frac{s+r}{rs}=\frac{r+s}{rs}$.
- This is the key move: the answer depends only on the sum $r+s$ and the product $rs$, so we never have to find the roots themselves.
💡 Adding two reciprocals stacks them into a single fraction whose top is the sum and bottom is the product.
7.EE.A.1 Step 3 Read off the sum and the product
- A quadratic with roots $r$ and $s$ can be written $a(x-r)(x-s)$.
- Expanding gives $a(x-r)(x-s)=a x^2 - a(r+s)x + a\,rs$.
- Matching this with $2003x^2 + 2004x + 2004$ (so $a=2003$) forces $-a(r+s)=2004$ and $a\,rs=2004$.
- Therefore $r+s=-\frac{2004}{2003}$ and $rs=\frac{2004}{2003}$.
💡 Expanding $(x-r)(x-s)$ shows the middle coefficient carries the sum and the constant carries the product, both scaled by the leading factor.
7.NS.A.2 Step 4 Divide to get the answer
- Substitute into $\frac{r+s}{rs}$: $\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}$.
- The top and bottom are the same size, $\frac{2004}{2003}$, differing only by the minus sign, so the quotient is $-1$.
- The $2003$ and $2004$ cancel out completely.
- The sum of the reciprocals of the roots is $-1$, which is choice (B).
💡 When numerator and denominator are equal in size but opposite in sign, their ratio is exactly $-1$.
8.EE.C.7 The $\frac{1}{x}$ term hides the fact that this is really a quadratic, so multip 6.EE.A.3 Call the two roots $r$ and $s$. The quantity we want is $\frac{1}{r}+\frac{1}{s} 7.EE.A.1 A quadratic with roots $r$ and $s$ can be written $a(x-r)(x-s)$. Expanding gives 7.NS.A.2 Substitute into $\frac{r+s}{rs}$: $\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}$ Review
Reasonableness: The clean cancellation is a sign the problem was built this way on purpose. For any quadratic $ax^2+bx+c=0$, the sum of the reciprocals of the roots is $\frac{r+s}{rs}=\frac{-b/a}{c/a}=-\frac{b}{c}$ — the leading coefficient $a$ never enters. Here $b=2004$ and $c=2004$, so the value is $-\frac{2004}{2004}=-1$ no matter what the leading $2003$ is. The awkward fraction $\frac{2003}{2004}$ 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.
Alternative: Substitute $y=\frac{1}{x}$ directly, so $x=\frac{1}{y}$. The equation becomes $\frac{2003}{2004}\cdot\frac{1}{y}+1+y=0$; multiplying by $y$ gives $y^2+y+\frac{2003}{2004}=0$. The roots of this new equation are exactly the reciprocals $\frac{1}{r}$ and $\frac{1}{s}$, and the sum of the roots of $y^2+y+\frac{2003}{2004}$ is $-\frac{1}{1}=-1$ — the answer, read off in a single step.
CCSS standards used (min grade 8)
8.EE.C.7Solve linear equations in one variable (Multiplying the equation through by $x$ and then by $2004$ to clear both fractions and produce $2003x^2+2004x+2004=0$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Combining $\frac{1}{r}+\frac{1}{s}$ into the single equivalent expression $\frac{r+s}{rs}$.)7.EE.A.1Apply properties of operations to add, subtract, factor, and expand linear expressions (Expanding $a(x-r)(x-s)$ and matching coefficients to read off $r+s=-\frac{2004}{2003}$ and $rs=\frac{2004}{2003}$.)7.NS.A.2Apply and extend understanding of multiplication and division of rational numbers (Dividing $-\frac{2004}{2003}$ by $\frac{2004}{2003}$ to get the final value $-1$.)
⭐ The sum of the reciprocals of a quadratic's roots is just $-\frac{b}{c}$ (the middle coefficient over the constant), so you can answer without ever finding the roots.
⭐ The sum of the reciprocals of a quadratic's roots is just $-\frac{b}{c}$ (the middle coefficient over the constant), so you can answer without ever finding the roots.
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