The quadratic equation $x^2+mx+n=0$ has roots that are twice those of $x^2+px+m=0,$ and none of $m,$ $n,$ and $p$ is zero. What is the value of $n/p?$
I want to find the value of $n/p$, so I need to relate $n$ and $p$ somehow. I notice that both equations are quadratic, so I can use the fact that the sum and product of the roots of a quadratic equation are related to the coefficients. Let me call the roots of the first equation $r_1$ and $r_2$, and the roots of the second equation $s_1$ and $s_2$. Then I know that $r_1+r_2=-m$, $r_1r_2=n$, $s_1+s_2=-p$, and $s_1s_2=m$. The problem also tells me that $r_1=2s_1$ and $r_2=2s_2$, since the roots of the first equation are twice those of the second equation. This means that $r_1+r_2=2(s_1+s_2)$, so I can substitute this into the equation for $m$ and get $-m=2(-p)$, or $m=2p$. Similarly, $r_1r_2=4(s_1s_2)$, so I can substitute this into the equation for $n$ and get $n=4m$. I get $n/p=4m/p=4(2p)/p=8$. 