The number of real solutions of $| x^{2} + 5 x + 4 | + 2 x + 6 = 0$ is

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Solution

$| x^{2} + 5 x + 4 | + 2 x + 6 = 0$

Case:1- $x^{2} + 5 x + 4 \geq 0$
$\Rightarrow (x + 1) (x + 4) \geq 0$
$\Rightarrow x \in (- \infty, - 4] \cup [- 1, \infty) . . . (1)$
The given equation becomes,
$x^{2} + 7 x + 10 = 0$
$\Rightarrow (x + 2) (x + 5) = 0$
$\Rightarrow x = - 2, - 5$
But, $x = - 2$ does not satisfy eqn(1)
$∴ x = - 5$ is the only solution in this case.

Case:2- $x^{2} + 5 x + 4 < 0$
$\Rightarrow (x + 1) (x + 4) < 0$
$\Rightarrow x \in (- 4, - 1) . . . (2)$
The given equation becomes,
$- (x^{2} + 5 x + 4) + 2 x + 6 = 0$
$\Rightarrow x^{2} + 3 x - 2 = 0$
$\Rightarrow x = \frac{- 3 \pm \sqrt{17}}{2}$
But, $x = \frac{- 3 + \sqrt{17}}{2}$ does not satisfy eqn(2)
$∴ x = \frac{- 3 - \sqrt{17}}{2}$ is the only solution in this case.

$\Rightarrow$ Only $2$ solutions.

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