How to solve $\left|\frac{x+4}{ax+2}\right| > \frac1x$ How to solve:
$$\left|\frac{x+4}{ax+2}\right| > \frac{1}{x}$$
What I have done:
I) $x < 0$:
Obviously this part of the inequation is $x\in(-\infty, 0), x\neq \frac{-2}{a}$
II) $x > 0$:
$$\left|\frac{x+4}{ax+2}\right| > \frac1x$$
$$\frac{|x+4|}{|ax+2|} > \frac1x$$
because $x > 0$ we can transform it to:
$x^2 + 4x > |ax + 2|$
$(x^2 + 4x)^2 > (ax + 2)^2$
$(x^2 + 4x)^2 - (ax + 2)^2 > 0$
$(x^2 + 4x - ax - 2) (x^2 +4x +ax + 2) > 0$
$(x^2 + (4-a)x - 2)(x^2 + (4+a)x + 2) > 0$
I am kind of stuck here. I also need to discuss the solution for various values of the parameter a. What is the easiest way out of this step? Something with the discriminants, perhaps?

The answer depends on the discriminants of the two quadratic polynomials:
$D_1= (a-4)^2+8$ and $D_2= (a+4)^2-8$. This shows that you have up to 4 roots.
Can you go on from here?
