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.

This is not a quadratic inequality but rather a reciprocal one, as we can shuffle the left-hand side a little:
$$3+\frac{4-x}x>0$$
$$3+\frac4x-1>0^*$$
$$2+\frac4x>0$$
$$\frac4x>-2$$
$$\frac1x>-\frac12$$
$$x<-2\text{ or }x>0$$
*Because we take $\frac xx=1$, we must exclude the $x=0$ case because it results in an indeterminate form.
