The sum of all possible value(s) of $a$ so that the equation $(x^{2} + 2 a x + 2 a + 3) (x^{2} + 2 a x + 4 a + 5) = 0$ has exactly three real and distinct roots is

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Solution

$(x^{2} + 2 a x + 2 a + 3) (x^{2} + 2 a x + 4 a + 5) = 0$
$D_{1} = 4 (a - 3) (a + 1); D_{2} = 4 (a + 1) (a - 5)$
For three real roots, different possibilities are as follows :

$(1) D_{1} = 0$ and $D_{2} > 0$
$D_{1} = 0 \Rightarrow a = 3, - 1$
and $D_{2} > 0 \Rightarrow a \in (- \infty, - 1) \cup (5, \infty)$
$∴$ No value of $a$ is possible.

$(2) D_{2} = 0$ and $D_{1} > 0$
$D_{2} = 0 \Rightarrow a = - 1, 5$
and $D_{1} > 0 \Rightarrow a \in (- \infty, - 1) \cup (3, \infty)$
$∴$ Only $a = 5$ is possible.

$(3) x^{2} + 2 a x + 2 a + 3 = 0$ and $x^{2} + 2 a x + 4 a + 5 = 0$ have exactly one root in common.
Let $α$ be their common root.
So $α^{2} + 2 α a + 2 a + 3 = 0$
And $α^{2} + 2 α a + 4 a + 5 = 0$
Subtracting the two, we get
$- 2 a - 2 = 0$
$\Rightarrow a = - 1$
But, for $a = - 1,$ both quadratic equations are identical.
So, $a$ can't be $- 1$

Hence, from above three cases, we get only one value of $a = 5$

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