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

Solution 1:

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 $f(a)=0$.

Solution 2:

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.

Solution 3:

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 $\beta=2\alpha$.