How Many *Integers* Satisfy

Let $f(x) = \dfrac {p(x)}{q(x)}$ be the LHS of the inequality. Then, we have:

$\begin{aligned} f(x) & = \frac {(x-3)^{\color{#D61F06}-\frac {|x|}x}\sqrt{(x-4)^2}(17-x)}{{\color{#3D99F6}\sqrt{-x}}(-x^2+x-1)(|x|-32)} & \small \color{#3D99F6} \sqrt{-x} \text{ is only defined for }x \le 0 \implies f(x) \text{ is defined for }x < 0. \\ & = \frac {(x-3)^{\color{#D61F06}1}|x-4|(17-x)}{\sqrt{-x}(-x^2+x-1)(|x|-32)} & \small \color{#D61F06} \text{For }x < 0 \implies - \frac {|x|}x = 1 \end{aligned}$

We note that, for $x<0$ , in the nominator $p(x)$ : $\begin{cases} x-3 < 0 \\ |x-4| > 0 \\ 17-x > 0 \end{cases}$ . Therefore, the nominator $p(x) < 0$ . And in the denominator $q(x)$ , for $x<0$ : $\begin{cases} \sqrt{-x} > 0 \\ -x^2+x-1 < 0 \end{cases}$ . Therefore, $f(x) < 0$ only if $|x|-32<0$ or $-31 \le x \le 1$ and there are $\boxed{31}$ integral $x$ satisfying the inequality.