The [quadratic equation](https://artofproblemsolving.com/wiki/index.php/Quadratic_equation) $x^2+mx+n$ has roots twice those of $x^2+px+m$, and none of $m,n,$ and $p$ is zero. What is the value of $n/p$?
$\textbf{(A) }\ {{{1}}} \qquad \textbf{(B) }\ {{{2}}} \qquad \textbf{(C) }\ {{{4}}} \qquad \textbf{(D) }\ {{{8}}} \qquad \textbf{(E) }\ {{{16}}}$
Solution 1
Let $x^2 + px + m = 0$ have roots $a$ and $b$. Then 
\[x^2 + px + m = (x-a)(x-b) = x^2 - (a+b)x + ab,\]
so $p = -(a+b)$ and $m = ab$. Also, $x^2 + mx + n = 0$ has roots $2a$ and $2b$, so
\[x^2 + mx + n = (x-2a)(x-2b) = x^2 - 2(a+b)x + 4ab,\]
and $m = -2(a+b)$ and $n = 4ab$. Thus $\frac{n}{p} = \frac{4ab}{-(a+b)} = \frac{4m}{\frac{m}{2}} = \boxed{\textbf{(D) }8}$. 
Indeed, consider the quadratics $x^2 + 8x + 16 = 0,\ x^2 + 16x + 64 = 0$.

Solution 2
If the roots of $x^2 + mx + n = 0$ are $2a$ and $2b$ and the roots of $x^2 + px + m = 0$ are $a$ and $b$, then using Vieta's formulas,
\[2a + 2b = -m\]
\[a + b = -p\]
\[2a(2b) = n\]
\[a(b) = m\]
Therefore, substituting the second equation into the first equation gives 
\[m = 2(p)\]
and substituting the fourth equation into the third equation gives
\[n = 4(m)\]
Therefore, $n = 8p$, so $\frac{n}{p}= \boxed{\textbf{(D) }8}$