2005 AMC 10B Problem 16

We need a link between a quadratic's coefficients and its roots. If a quadratic $x^2 + Bx + C$ has roots $r$ and $s$, then it equals $(x - r)(x - s) = x^2 - (r + s)x + rs$. So
- the sum of the roots is $r + s = -B$, and
- the product of the roots is $rs = C$.

1. First quadratic, $x^2 + px + m$. Call its roots $r$ and $s$. Then $r + s = -p$ and $rs = m$.
2. Second quadratic, $x^2 + mx + n$. Its roots are $2r$ and $2s$. Then
- sum: $2r + 2s = -m$,
- product: $(2r)(2s) = 4rs = n$.
3. Compare the sums. $2r + 2s = 2(r + s) = 2(-p) = -2p$. This also equals $-m$, so $m = 2p$.
4. Compare the products. $n = 4rs = 4m$.
5. Put them together. $n = 4m = 4 \cdot 2p = 8p$. Since $p$ is not zero, we can divide: $\frac{n}{p} = 8$.

Check with real numbers: $x^2 + 8x + 16 = (x + 4)^2$ has root $-4$ (twice), so $p = 8$, $m = 16$. Doubling the root gives $-8$, and $(x + 8)^2 = x^2 + 16x + 64$, which has middle coefficient $16 = m$ and $n = 64$. Indeed $64 / 8 = 8$.

The answer is $\boxed{\textbf{(D)}\ 8}$.

Any question about roots can be turned into a question about their sum and product (these are called Vieta's formulas). Doubling every root doubles the sum but multiplies the product by $2 \cdot 2 = 4$. The letter $m$ shows up in both quadratics, and it is the bridge that connects $n$ to $p$.

Call the roots of $x^2 + px + m$ by $r$ and $s$. Vieta's formulas give
$$r + s = -p, \qquad rs = m.$$

The other quadratic, $x^2 + mx + n$, has roots $2r$ and $2s$, so
$$2r + 2s = -m, \qquad (2r)(2s) = n.$$

Compare: the sum equation says $-m = 2(r+s) = -2p$, so $m = 2p$. The product equation says $n = 4rs = 4m$.

Chain them: $n = 4m = 4(2p) = 8p$, and since $p \ne 0$,
$$\frac{n}{p} = 8.$$

The answer is $\boxed{\textbf{(D)}\ 8}$.