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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 $m=3$.

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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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    $\begingroup$ SO I was proceeding correctly, but didn't have smart to go. Thanks a lot! $\endgroup$ – gilly Nov 17 '17 at 10:24

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