absolute fraction of polynomials I'm trying to solve inequality:
$$\bigg|\frac{x^2+2x-36}{x^2-4}\bigg|\gt 1$$
The first step I do is to conclude how the function behaves (to see when it's negative):
$$\frac{x^2+2x-36}{x^2-4} = 0$$
My approach is to solve the nominator and denominator separately (as separate second degrees polynomials). The denominator is easy (opens upward parabola):
$$x^2-4=0 \\(x-2)(x+2)=0\iff x=-2 \vee x=2
$$
The issue I have is with the nominator:
$$x^2+2x-36 = 0$$
Since the standard approach with quadratic formulas doesn't look sensible to me.
Could you share some hints? How I should approach that?
 A: Consider the equality that lies at the boundary of the inequality.
$$\bigg|\frac{x^2+2x-36}{x^2-4}\bigg| = 1$$
$$\frac{x^2+2x-36}{x^2-4} = \pm 1$$
$$x^2+2x-36 = \pm ({x^2-4})$$
$$x^2+2x-36 = x^2-4 \lor x^2+2x-36 = -x^2+4$$
$$2x-32 = 0 \lor 2x^2+2x-40 = 0$$
$$x = 16 \lor (x = 4 \lor x = -5)$$
This gives you three “special” values of $x$ to work with, in addition to the two you already found $x = \pm 2$, where the denominator is 0.  In order, these are:

*

*$x = -5$ (Polynomial fraction is -1.)

*$x = -2$ (Polynomial fraction is undefined.)

*$x = 2$ (Polynomial fraction is undefined.)

*$x = 4$ (Polynomial fraction is -1.)

*$x = 16$ (Polynomial fraction is 1.)

These five values partition the real line into six intervals:

*

*$x \in (-\infty, -5)$

*$x \in (-5, -2)$

*$x \in (-2, 2)$

*$x \in (2, 4)$

*$x \in (4, 16)$

*$x \in (16, \infty)$
Now, you just have to go through these six cases and determine which ones satisfy your original inequality.
A: $$
\begin{aligned}
& \left|\frac{x^2+2 x-36}{x^2-4}\right|>1, \text { where } x \neq \pm 2 \\
\Rightarrow \quad& \left|x^2+2 x-36\right|>\left|x^2-4\right| \\
\Rightarrow \quad&\left(x^2+2 x-36\right)^2>\left(x^2-4\right)^2 \\
\Rightarrow \quad& \left(x^2+2 x-36\right)^2-\left(x^2-4\right)^2>0 \\
\Rightarrow \quad& \left(x^2+2 x-36+x^2-4\right)\left(x^2+2 x-36-x^2+4\right)>0 \\
\Rightarrow \quad& 4\left(x^2+x-20\right)(x-16)>0 \\
\Rightarrow \quad& 4(x+5)(x-4)(x-16)>0 \\
\Rightarrow \quad& -5<x<4 \text { or } x>16
\end{aligned}
$$
However $x\ne \pm 2$, therefore the solutions  are
$$\boxed{-5<x<-2, \quad-2<x<2, \quad 2<x<4 \quad\textrm{  and  }x>16.}$$
