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.
 A: We have
$$3+\frac{4-x}{x} >0$$
$$\frac{3x+4-x}{x} >0$$
$$\frac{2x+4}{x} >0$$
Divide both sides by $2$
$$\frac{x+2}{x} >0$$
Now use wavy curve method
$-\infty+++++ (-2)-----(0)++++++ \infty$ 
As we want L.H.S. to be positive, hence we get
$x \in (-\infty, -2) \cup (0, \infty)$
Edit 1
Let me explain you how I decided the signs.
When $x>0$, then $x+2>0$ and $x>0$, hence $\frac{x+2}{x}$ will be positive.
Again when $x<-2$, then $x+2<0$ and $x<0$, hence $\frac{x+2}{x}$ will be positive.
When $-2<x<0$, then $x+2>0$ but $x<0$, hence $\frac{x+2}{x}$ will be negative.
That is how I decided the sign.
A: 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.
A: 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
$$L = \{x>0, x>-2\} \cup \{x<0,x<-2\} = \{x>0\} \cup \{x<-2\} = (-\infty,-2) \cup (0, \infty).$$
A: We can use the following

$\frac{A}{B}>0$ and $B\not=0$ if and only if $AB>0$.

It holds for the other remaining inequality signs ($<$, $\leq$, etc). Trivially, it also holds for $=$.
So after obtaining the following,
$$\frac{x+2}{x}>0$$
where $x\not=0$, we can rewrite it as
$$x(x+2)>0$$
As the other answer called it as "wavy curve" method, draw a number line with the zeros of $x(x+2)$ and determine the sign of $x(x+2)$ at each interval.
Finally select the intervals with positive signs (because of $>$ in $\frac{x+2}{x}>0$).
Thus the solution is
$$x<-2\text{ or }x>0$$
A: Hint:
I suppose that your inequality is:
$$
3+\frac{4-x}{x}>0
$$
that, for $x\ne 0$ becomes
$$
\frac{2x+4}{x}>0
$$
now note that a fraction is positive if the numerator and the denominator have the same sign.

The inequality is equivalent to the two  systems:
$$
\begin{cases}
2x+4>0\\
x>0
\end{cases}
\quad \mbox{or} \quad
\begin{cases}
2x+4<0\\
x<0
\end{cases}
$$
with solution:
$x<-2$ or $x>0$
