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I've been stuck on this problem for quite a while even though it seemed trivial to me at first.

Basically, I have this:

$$\lvert ax+4 \rvert>\frac1x$$

It is quite easy to conclude that only $x>0$ is possible. For the case when $a=0$, it's trivial to solve too.

But other than that, I've been stuck for a while at it. Any help would be appreciated.

Thanks!

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Infact clearly $x< 0$ is always a solution. Positive solutions must satisfy (by squaring) $$x^2(ax+4)^2>1 \implies (ax^2+4x+1)(ax^2+4x-1)> 0$$ Given $a$, you could factor the quadratics further and find points where the LHS can change sign (which are the roots of the LHS). Rest is checking signs of the expression between those roots.

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