2025 AMC 10A Problem 18

The definition asks only for the reciprocals of the roots, so there is no need to find a single root.

First count the roots. For each $k$ from $1$ to $2025$ the factor $kx^2 - 4x - 3$ has discriminant $16 + 12k > 0$, so it contributes two distinct real roots. No two factors share a root: if $kx^2 - 4x - 3 = 0$ and $jx^2 - 4x - 3 = 0$ with $k \ne j$, subtracting gives $(k-j)x^2 = 0$, so $x = 0$, but $x = 0$ makes each factor $-3$. Hence there are exactly $2 \cdot 2025 = 4050$ real roots, matching the degree.

Now take the two roots $p, q$ of the $k$th factor. Vieta gives
$$p + q = \frac{4}{k}, \qquad pq = -\frac{3}{k},$$
so
$$\frac1p + \frac1q = \frac{p+q}{pq} = \frac{4/k}{-3/k} = -\frac43,$$
the same value for every $k$. The $k$ cancels, which is the whole point of the construction.

Summing over all $2025$ factors, the reciprocals of the $4050$ roots total
$$2025 \cdot \left(-\frac43\right) = -2700.$$
Their arithmetic mean is $\dfrac{-2700}{4050} = -\dfrac{2}{3}$, and the harmonic mean is its reciprocal, $-\dfrac{3}{2}$.

The answer is $\boxed{\textbf{(B)}\ -\frac{3}{2}}$.

A harmonic mean is an arithmetic mean in disguise: it lives entirely in the world of reciprocals. Whenever the numbers involved are roots of polynomials, the sum of reciprocals is the Vieta ratio $-\frac{\text{linear coefficient}}{\text{constant term}}$, which is often far easier to read off than the roots themselves. The same trick handles any product of quadratics whose linear and constant coefficients are fixed while the leading coefficient varies: the per-factor reciprocal sum is constant, so the answer does not depend on how many factors there are.

The reciprocals of the roots of a polynomial are the roots of its reverse. Reversing $kx^2 - 4x - 3$ gives $-3y^2 - 4y + k$, i.e. $3y^2 + 4y - k = 0$, whose two roots sum to $-\frac43$ by Vieta, again independently of $k$. Multiplying the $2025$ reversed quadratics produces a degree-$4050$ polynomial whose roots are precisely the $4050$ reciprocals, and their total is $2025 \cdot \left(-\frac43\right) = -2700$ as before.

• Dividing the sum of reciprocals by 2025 (the number of factors) rather than 4050 (the number of roots), which turns the answer into -3/4 and tempts a guess at the nearest choice.
• Taking the reciprocal only once, and reporting the mean of the reciprocals $-\frac{2}{3}$, which is choice (E).
• Worrying about whether some factor has complex roots; the discriminant $16 + 12k$ is positive for every $k \ge 1$, so all $4050$ roots are real and the "real roots" wording changes nothing.