Question

1. $|x^{2}-3x - 18|>0$

Explanation:

Step1: Factor the quadratic inside absolute - value

Factor $x^{2}-3x - 18$ as $(x - 6)(x+3)$. So we have $|(x - 6)(x + 3)|>0$.

Step2: Analyze when the absolute - value is zero

The absolute - value $|(x - 6)(x + 3)| = 0$ when $(x - 6)(x + 3)=0$. Solving $(x - 6)(x + 3)=0$ gives $x = 6$ or $x=-3$.

Step3: Determine the solution set

Since we want $|(x - 6)(x + 3)|>0$, the solution set is all real numbers except $x = 6$ and $x=-3$. In interval notation, the solution is $(-\infty,-3)\cup(-3,6)\cup(6,\infty)$.

Answer:

$x\in(-\infty,-3)\cup(-3,6)\cup(6,\infty)$