How to solve this inequality question without manual checking?

Question

Question:

Find the maximum integral value which satisfies: $$\frac{x-2}{x^2-9}<0$$

I know that this means either of the following:

#1. $x-2<0$ and $x^2-9>0$. Implies that $x \in (3, 2)$, which is false. ^[1]

#2. $x-2>0$ and $x^2-9<0$. Implies that $x \in (2, 3)$, which is true. ^[1]

However, this does not give the correct answer. Of course, I can solve this by checking values, but I want a proper method of solving.

I, thus, need help in figuring out the method to get the correct answer

^[1] - $x^2-9>0 \Rightarrow x^2>9 \Rightarrow x>3$

Answer 1

What about "standard" approach to solving such inequalities in real variable $x$? $$ \frac{x-2}{x^2-9} < 0\qquad\qquad |x|\neq3 $$ Now multiply both sides by $(x^2-9)^2$ which is always positive and does not change the inequality sign. $$ (x-2)(x^2-9) < 0 $$ $$ (x-2)(x-3)(x+3) < 0 $$ We got a fairly easy polynomial inequality. Given $|x|\neq3$ the solution is: $$ x\in (-\infty,-3)\cup(2,3) $$ The maximum integer value from this set is obviously $-4$.

Answer 2

Because of the denominator, $|x|=3$ is forbidden. If $|x|<3$, we need $x-2>0$, which is impossible for integer $x$. If $|x|>3$, we need $x-2<0$, which together with $|x|>3$ implies that $x\le -4$ (integer!), so $x=-4$ is the answer.

Answer 3

To solve this quick, look at your function $\frac{x-2}{x^2-9}<0$ and find all the roots of its polynomial. Here it is very easy: it equals to $\frac{x-2}{(x-3)(x+3)}<0$

So its numerator/denominator zeroes go in the following order: $-3, 2, 3$. At each point exactly one multiplier changes sign, so the sign of the whole expression changes to the opposite.

As there is no square here, this means that the function's value is negative before -3 (where it does not exist), then positive before 2, crosses zero in 2, goes negative until 3, then it turns positive again. So the inequality is true in $(-\infty,-3) \cup (2,3) $, and in here the maximum integer value is, obviously, 4.

Look here for the method. http://www.bymath.com/studyguide/alg/sec/alg28.html . That's someting we used back at school.