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

Let the roots be $\alpha, 2\alpha$.
By Vieta's formulas,
Sum of roots $=3\alpha = 5m$, product of roots $=2\alpha^2 = 10m-4$.
Hence, $2(\frac{5m}3)^2 = 10m-4 \\ \implies 25m^2 - 45m +18 = 0$, we get (by simple factorisation) $m=0.6$ or $1.2$.
Test by plugging these back into the original expression and getting $f(x) = x^2-3x+2$ and $f(x) = x^2 - 6x+8$ respectively, which each have two distinct roots with one twice the other.
