Solving quadratic inequalities I have a quadratic inequality I am halfway through solving but cannot quite figure out the concept.
$$3+\frac{4-x}x>0$$
I so far understand that without knowing whether x is positive, I would need to square it, but I do not know what comes next.

We want to find all Solutions to $3 + \frac{4-x}{x} >0$.
At first, we have to make a distinction between (i) $x>0$ and (ii) $x<0$. Since $x=0$ is not possible, all possible values of $x$ are covered by this distinction.
(i) We multiply the inequality by $x$. Since $x$ is positive, the ">" remains a ">":
$$3x + 4 - x > 0 \Longleftrightarrow 2x > -4 \Longleftrightarrow x>-2.$$
(ii) We multiply the inequality by $x$. Since $x$ is negative, this time the ">" changes to a "<":
$$3x + 4 - x < 0 \Longleftrightarrow 2x < -4 \Longleftrightarrow x<-2.$$
Conclusion: The set "L" of all solutions can be written as the Union
$$L = \{x>0, x>-2\} \cup \{x<0,x<-2\} = \{x>0\} \cup \{x<-2\} = (-\infty,-2) \cup (0, \infty).$$
