Number of integral values of x satisfying the inequality~ x^ 2 + 6x...

We have& $\dfrac{x^{2} + 6x - 7}{\left|x + 2\right|\left|x + 3\right|} < 0$.

Clearly by modulus function~property~\left|x + 2\right| \ge 0 \text{ and } \left|x + 3\right| \ge 0$~and~$x \ne - 2, - 3$.</p><p>Now,$x^{2} + 6x - 7 < 0$.</p><p>$⇒ x^{2} + 7x - x - 7 < 0$</p><p>$⇒ \left(x + 7\right)\left(x - 1\right) < 0$</p><p>Using Sign Scheme for this expression,</p><p>$⇒ x \in \left(- 7, 1\right) - \left\{- 3, - 2\right\}$</p><p>So, integers are \left\{- 6, - 5, - 4, - 1, 0\right\}$.

So, total number of integral~values of~$x = 5$.