Solve : $∣ ∣ x^{2} + 4 x + 3 ∣ ∣ + 2 x + 5 = 0$ . The number of integral solution/s is/are

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Solution

$∣ ∣ x^{2} + 4 x + 3 ∣ ∣ + 2 x + 5 = 0$
$| (x + 1) (x + 3) | + 2 x + 5 = 0$
Case I:- $x \leq - 3$ and $x \geq - 1$
$x^{2} + 4 x + 3 + 2 x + 5 = 0$
$x^{2} + 6 x + 8 = 0$
which gives $x = - 4, - 2$ , but $x = - 2$ doesnot belong to the considered range
So, $x = - 4$ only satisfies this case.
Case II:- $- 3 \leq x \leq - 1$
$x^{2} + 4 x + 3 - 2 x - 5 = 0$
$x^{2} + 2 x - 2 = 0$
$x = \frac{- 2 \pm \sqrt{12}}{2} = - 1 \pm \sqrt{3}$
No integral value of $x$ satisfies this case.

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