The sum of all real values of $^{'} m^{'}$ such that the equation $(x^{2} - 2 m x - 4 (m^{2} + 1)) . (x^{2} - 4 x - 2 m (m^{2} + 1))$ has exactly three different real root is.

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Solution

Let $f_{1} (x) = x^{2} - 2 m x - 4 (m^{2} + 1)$
$D_{1} = \sqrt{{(- 2 m)}^{2} - 4 (- 4 (m^{2} + 1))} = \sqrt{{4 m}^{2} + {16 m}^{2} + 4}$
$D_{1} = \sqrt{{20 m}^{2} + 4} \neq 0$
$f_{2} (x) = x^{2} - 4 x - 2 m (m^{2} + 1)$
$D_{2} = \sqrt{{(- 4)}^{2} - 4 (- 2 m (m^{2} + 1))} = \sqrt{16 + {8 m}^{3} + 8 m}$
Now, for end roots, $\sqrt{D_{2}} = 0$
So, $\sqrt{16 + {8 m}^{3} + 8 m} = 0 \Rightarrow {8 m}^{3} + 8 m + 16 = 0$
$m^{2} + m + 2 = 0 \Rightarrow (m + 1) (m^{2} - m + 2) = 0$
So, $m = - 1$ , $\frac{1}{2} + \frac{\sqrt{7}}{2} i$ , Thus sum of all value of $1 + 0.5 + 0.5 = 2$

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