Explanation

(b)
We have,
$\left| x^{2} + 4x + 3 \right| + 2x + 5 = 0$
Here two cases arise.
Case I When $x^{2} + 4x + 3 > 0$
$\Rightarrow \ x^{2} + 4x + 3 + 2x + 5 = 0$
$\Rightarrow x^{2} + 6x + 8 = 0$
$\Rightarrow (x + 2)(x + 4) = 0$
$\Rightarrow x = - 2,\ - 4$
$x = - 2$ is not satisfying the condition $x^{2} + 4x + 3 > 0$. So $x = - 4$ is the only solution of the given equation.
Case II When $x^{2} + 4x + 3 < 0$
$\Rightarrow \ - {(x}^{2} + 4x + 3) + 2x + 5 = 0$
$\Rightarrow \ - x^{2} - 2x + 2 = 0$
$\Rightarrow x^{2} + 2x - 2 = 0$
$\Rightarrow \left( x + 1 + \sqrt{3} \right)\left( x + 1 - \sqrt{3} \right) = 0$
$\Rightarrow x = - 1 + \sqrt{3}, - 1 - \sqrt{3}$
Hence, $x = - (1 + \sqrt{3})$ satisfy the given condition.
Since, $x^{2} + 4x + 3 < 0$ while $x = - 1 + \sqrt{3}$ is not satisfying the condition. Thus, number of real solutions are two.