Quadratic Duos

There are three cases:

Case I: the first quadratic has two equal roots and the second has two different roots;

Case II: the first quadratic has two different roots and the second has two equal roots;

Case III: the two quadratics has one common root.

Case I: the first quadratic has two equal roots and the second has two different roots

Let the two equal roots of quadratic $x^{2}-2mx-4(m^{2}+1)$ be $p$ .

From Vieta's Formulas,

$2p = 2m$

$p^{2} = -4(m^{2}+1)$

Then $m^{2} = -4(m^{2}+1)$

$5m^{2} = -4$

But $m$ is a real number, so there is no $m$ satisfying this case.

Case II: the first quadratic has two different roots and the second has two equal roots

Let the two equal roots of quadratic $x^{2}-4x-2m(m^{2}+1)$ be $q$ .

From Vieta's Formulas,

$2q = 4$

$q^{2} = -2m(m^{2}+1)$

Then $m^{3}+m+2 = 0$

$(m+1)(m^{2}-m+2) = 0$

The polynomial $m^{2}-m+2$ has two non-real roots.

If $m = -1$ , then the given polynomial becomes

$(x^{2}+2x-8)(x^{2}-4x+4)$

$(x-2)(x+4)(x-2)(x-2)$

There are only two different roots for this case, so $m$ cannot equal $-1$ .

Case III: the two quadratics has one common root

Let the roots of quadratic $x^{2}-2mx-4(m^{2}+1)$ be $a$ and $b$ ;

and the roots of quadratic $x^{2}-4x-2m(m^{2}+1)$ be $a$ and $c$ ;

with $a \ne b \ne c$ .

So that the given equation has three different roots: $a$ , $b$ , $c$ .

From Vieta's Formulas,

(1): $a+b = 2m$

(2): $ab = -4(m^{2}+1)$

(3): $a+c = 4$

(4): $ac = -2m(m^{2}+1)$

(5): (1)-(3): $b-c = 2m-4$

(6): (2)-(4): $a(b-c) = (m^{2}+1)(2m-4)$

But $b \ne c$ , so $2m-4 \ne 0$

(7): (6)/(5): $a = m^{2}+1$

From (2) and (7): $b = -4$

From (1): $m^{2}-3 = 2m$

$m^{2}-2m-3 = 0$

$(m-3)(m+1) = 0$

$m = 3, -1$

If $m = 3$ , then $a = 10$ , $b = -4$ and $c = 4-a = -6$ .

If $m = -1$ , then $a = 2$ , $b = -4$ and $c = 4-a = 2$ .

But the three roots must be different, so $m$ cannot equal $-1$ .

Therefore $m = 3$ , and the sum of all possible values of $m$ is $3$ .