Solving the inequality $(x^2+3)/x\le 4$ This is the inequality
$$\left(\frac {x^2 + 3}{x}\right) \le 4 $$
This is how I solve it
The $x$ in the left side is canceled and $4x$ is subtracted from both sides.
$$\not{x} \left (\frac {x^2+3} {\not{x}}\right) \le 4x $$
$$ x^2+3 - 4x \le 4x - 4x $$
$$x^2 -4x + 3 \le 0 $$
Then the trinomial is factorized
$$(x-3)(x-1) \le 0 $$
Therefore
$$1\lt x \lt 3 $$
Can someone tell me if I make a mistake, or if the process is good enough. Will you have done something different?
 A: Hints:
$$\frac{x^2+3}x\le 4\implies\frac{x^2-4x+3}x\le 0\iff \frac{(x-3)(x-1)}x\le 0\;\;(*)$$
and now you can apply what we call here "the snake method": on the real line, mark the points where the expressions numerator or denominator vanish, which are $\;0,1,3\;$ , and within one of these domains, namely $\;x< 0\;,\;\;0<x\le 1\;,\;\;1\le x\le 2\;$ find the sign of the expression...and then the signs alternate from domain to domain!
For example, here we easily find that at $\;x=2\;$ ( i.e., in $\;1< x\le3\;$), the expression's sign is minus, so the domains signs are
$$\begin{align}&x<0&||&\;\;\;0<x\le 1&||&\;\;\;1\le x\le 3&||&\;\;\;3\le x\\{}\\
&\;\;\color{red}{--}&||&\;\;\;\;\;\;++&||&\;\;\;\;\;\;\color{red}{--}&||&\;\;\;\;++\end{align}$$
and now just choose the correct ones for what you want:
$$\color{green}{x<0\;\;or\;\;1\le x\le3}$$
Midly interesting exercise: why and how the above works? Pay attention to the fact that all the factors in the equality (*) have odd power (in this case, $\;1\;$)
A: Never multiply an inequality by a variable since when multiplying by a negative number you have to change the sign.This is how this kind of problems should be solved
$$\frac{x^2+3}{x}\leq 4\\\frac{x^2+3}{x}-4\leq 0\\\frac{x^2-4x+3}{x}\leq 0\\\frac{(x-3)(x-1)}{x}\leq 0$$
Now you should spit it into cases


*

*$x<0$ Then you have that $x,(x-1),(x-3)$ are all negative so everything is negative

*$0<x<1$ Then you have that $x$ is positive and $(x-1),(x-3)$ negative hence the expression is positive since 2 negatives give positive

*$1<x<3$ Then you have that $x,(x-1)$ are positive and $x-3$ is negative hence the expression is negative

*$x>3$ everything is positive hence expression is positive


Now you see that $x<0$ and $1<x<3$ are solutions,now to check the boundaries,$x=0,1,3$ you can see that for $x=0$ it's undefined and that $x=1,3$ fit so the solution is $x\in(-\infty,0)\cup [1,3]$
A: To deal away with fractions, we can multiply both sides of $\left(\frac {x^2 + 3}{x}\right) \le 4$ with $x^2>0$ (here, $x\neq 0$):
$$
(x^2+3)x\leq 4x^2\iff x^3+3x-4x^2\leq 0\iff x(x-1)(x-3)\leq 0.
$$
We want the product of 3 factors above to be nonpositive so there is an odd number of nonnegative factors among $x-3<x-1<x$. This can only happen when 
$$
x-3<x-1<x<0 \quad\text{or}\quad x-3\leq 0\leq x-1<x.
$$
Thus, the answers are $x<0$ or $1\leq x\leq3$.
A: The problem is that when you multiply an inequality by a negative number, the inequality reverses.  What you have started is the case where $x$ is positive.  The entire interval you get is positive, so the entire interval works.
The negative case, on the other hand, is
$$(x-3)(x-1)\ge0$$
$$x<1\text{ or }x>3$$
The only cases here where $x<0$, is, well... $x<0$.  You will notice from your original equation that for negative $x$, the left side is negative and clearly less than $4$.
