We wish the roots of the following quartic to be real and and distinct, but two roots should be equal. Eg. Roots should be $a,b,c,c$ where $a,b,c$ are Real and distinct.
We have to find values of $m$ corresponding to this condition.
I observed that discriminant of first quadratic is positive, and discriminant of second quadratic is $0$ at $m=-1$. BUT when $m=-1$, then the two quadratic have a common root! So the roots are $-4,1,1,1$. This is not required.
Now I think if we find common root by subtracting quadratics, then we get desired value of $m$. On subtracting quadratic I got:
$$(m-2)x = (m^2+1)(m-2)$$
Meaning either $m=2$ or $x=m^2+1$. Still none lead to answer.
The answer is $m=3$.