<s>[INST] 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
 [/INST] The given equation can be written as
$\begin{align}
f(x)& =x^2-4-5mx+10m\\
&=(x-2) (x+2) -5m(x-2) \\
&=(x-2) (x+2-5m) 
\end{align}$
Now, we have two possibilities: $\alpha=2\beta$ or $\beta=2\alpha$.
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