Question

If the roots of $x^2+mx+n=0$ are twice the roots of $x^2+px+m=0,p\ne 0$, then the value of $\frac{n}{p}$ is

Answer 1

Let the roots of $x^2+px+m=0$ be $\alpha ,\beta$
$\alpha +\beta =-p$ and $\alpha \beta =m\cdots \left(1\right)$

Roots of $x^2+mx+n=0$ are $2\alpha ,2\beta$
$\Rightarrow 2\alpha +2\beta =-m$ and $2\alpha \cdot 2\beta =n$
$\Rightarrow 2(\alpha +\beta )=-m$ and $4\alpha \beta =n$
$\Rightarrow 2p=m$ and $4m=n\left[\text{Using}\right(1\left)\right]$
$\Rightarrow 8p=n$
$\Rightarrow \frac{n}{p}=8$