<s>[INST] 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?
 [/INST] 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?
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