2025 AMC 10A Problem 18

The harmonic mean is built from the reciprocals of the roots, so what we really need is the sum of those reciprocals. We never need the roots themselves.

Step 1: try the first two factors.

1. $k = 1$: $x^2 - 4x - 3 = 0$ has roots $2 + \sqrt7$ and $2 - \sqrt7$. Their sum is $4$ and their product is $4 - 7 = -3$. So
$$\frac{1}{2 + \sqrt7} + \frac{1}{2 - \sqrt7} = \frac{\text{sum}}{\text{product}} = \frac{4}{-3} = -\frac43.$$
2. $k = 2$: $2x^2 - 4x - 3 = 0$. By Vieta's formulas, the roots add to $\frac{4}{2} = 2$ and multiply to $-\frac{3}{2}$. So the reciprocals add to $\frac{2}{-3/2} = -\frac43$.

The same number both times. That is worth checking in general.

Step 2: every factor gives $-\frac43$. For $kx^2 - 4x - 3 = 0$, Vieta's formulas say the two roots $p$ and $q$ satisfy
$$p + q = \frac{4}{k}, \qquad pq = -\frac{3}{k}.$$
Adding the reciprocals over a common denominator:
$$\frac1p + \frac1q = \frac{p + q}{pq} = \frac{4/k}{-3/k} = -\frac43.$$
The $k$ cancels, so every factor gives $-\frac43$.

Step 3: make sure there really are $4050$ real roots.

1. Each factor has discriminant $(-4)^2 - 4 \cdot k \cdot (-3) = 16 + 12k$, which is positive. So each factor has two different real roots.
2. Two different factors never share a root. If $x$ solved both $kx^2 - 4x - 3 = 0$ and $jx^2 - 4x - 3 = 0$, subtracting would give $(k - j)x^2 = 0$, so $x = 0$. But $x = 0$ makes every factor equal $-3$, not $0$.
3. So there are $2 \cdot 2025 = 4050$ different real roots.

Step 4: compute the harmonic mean.

1. Sum of all the reciprocals: $2025 \cdot \left(-\frac43\right) = -2700$.
2. Average of the reciprocals: divide by the number of roots, $4050$: $\frac{-2700}{4050} = -\frac23$.
3. The harmonic mean is the reciprocal of that average: $-\frac32$.

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

A harmonic mean is just an ordinary average taken in the world of reciprocals, and then flipped back. For the roots of a quadratic, the reciprocals add to $\frac{\text{sum of roots}}{\text{product of roots}}$, which Vieta's formulas give straight from the coefficients. Here only the $x^2$ coefficient changes from factor to factor, and it cancels, so every factor contributes the same amount. When a problem gives a huge product like this, try the first one or two cases to spot what stays the same.

Replace $x$ by $\frac1y$, so that $y$ stands for the reciprocal of a root. The equation $kx^2 - 4x - 3 = 0$ becomes
$$\frac{k}{y^2} - \frac{4}{y} - 3 = 0.$$
Multiply by $y^2$: $k - 4y - 3y^2 = 0$, or $3y^2 + 4y - k = 0$. The two roots of this equation are exactly the reciprocals of the two roots of the original factor, and by Vieta's formulas they add to $-\frac43$, whatever $k$ is. The rest is the same: the $4050$ reciprocals add to $-2700$, their average is $-\frac23$, and the harmonic mean is $-\frac32$.