<s>[INST] Condititon for roots of quartic to be real and two be coincident 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.
$$[x^2-2mx-4(m^2+1)][x^2-4x-2m(m^2+1)]$$
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 $m=3$.
 [/INST] Let $f(x)=x^2-2mx-4(m^2+1),g(x)=x^2-4x-2m(m^2+1)$.
Also, let $c\in\mathbb R$ be the double root of $f(x)g(x)$.
We have three cases to consider : 
Case 1 : $x=c$ is a double root of $f(x)$
Case 2 : $x=c$ is a double root of $g(x)$
Case 3 : $f(c)=g(c)=0$


*

*Case 1 : If $x=c$ is a double root of $f(x)$, then we have to have $(-2m)^2-4\times 1\times (-4(m^2+1))=0$, but there are no such $m\in\mathbb R$.

*Case 2 : If $x=c$ is a double root of $g(x)$, then solving $(-4)^2-4\times 1\times (-2m(m^2+1))=0$ gives $m=-1$. Then, we have $f(x)=(x+4)(x-2),g(x)=(x-2)^2$ which don't satisfy our condition.

*Case 3 : If $f(c)=g(c)=0$, then from $0=f(c)-g(c)=-2(m-2)(c-m^2-1)$, we have $m=2$ or $c=m^2+1$. If $m=2$, then $f(x)=g(x)$ which don't satisfy our condition. If $c=m^2+1$, then by Vieta's formulas, $x=-4$ is a root of $f(x)$, so $f(-4)=0\implies m=-1,3$. We already see that $m\not=-1$. If $m=3$, then we have $f(x)=(x-10)(x+4),g(x)=(x-10)(x+6)$ which satisfy our condition.
Therefore, the answer is $\color{red}{m=3}$.
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