Find all possible values $m$ such that $f(x)=x^2-5mx+10m-4$ has 2 roots and one of them is twice as the other Consider the polynomial $f(x)=x^2-5mx+10m-4$, find $m$ such that there exist a number $a$ that satisfies $f(a)=f(2a)=0$. This was my attempt:
$f(a)=f(2a)$ 
$a^2-5ma+10m-4=4a^2-10ma+10m-4$ 
$a^2-5ma=4a^2-10ma$ 
$-3a^2+5ma=0$ 
$a*(-3a+5m)=0$
If $a=0$ then $f(a)=10m-4=0$ and $m=\frac{2}{5}$, else if $-3+5m=0$ then I can't solve.
Is my first answer correct? And if yes, how do I solve for the second one

Your equation always has two roots: $2$ and $5m-2$. So, one of them is twice the other if and only if $m=\frac65$ or if $m=\frac35$.
In your approach, it seems that you forgot that you should also have $f(a)=0$.
