<s>[INST] Solving $\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4}$. Solve the inequality
$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow$ $\frac{x^2-2}{x^2+2}  - \frac{x}{x+4} \leq 0 \Rightarrow$ $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0 \Rightarrow$ $x \geq \frac{1}{2} \pm \sqrt{2.25} \Rightarrow$ $x_1 \geq -1, \; x_2 \geq 2$ 
We notice that $x_2 \geq 2$ is a false root and testing implies that the solutions of the inequality lies within the interval $-1 \leq x \leq 2$.
Problem: But $x<-4$ also solves the inequality, so I must have omitted or done something wrong? And also, am I using implication and equivalence symbols correctly when doing the calculations?
Thank you for your help!
 [/INST] You incorrectly arrive at 
$$
\frac{4(x^2-x-2)}{(x^2+2)(x+4)}\ge0
$$
which should be instead
$$
\frac{4(x^2-x-2)}{(x^2+2)(x+4)}\le0
$$
Moreover you do the “illegal” step of removing the denominator, which changes sign at $-4$.
The factors $4$ and $x^2+2$ can be removed because they're positive. Factoring the numerator we obtain
$$
\frac{(x+1)(x-2)}{x+4}\le0
$$
that's the same as
$$
\frac{(x+4)(x+1)(x-2)}{(x+4)^2}\le0
$$
The denominator is positive, so it's irrelevant as far the inequality is concerned, but we have to remember that $-4$ is a value to exclude from solutions, because it makes the expression undefined.
Thus we are reduced to
$$
(x+4)(x+1)(x-2)\le 0\qquad x\ne-4
$$
A polynomial only changes sign at odd multiplicity roots. For large values of $x$ the polynomial $(x+4)(x+1)(x-2)$ is positive. Going backwards on the real line it changes sign at $2$ (becoming negative), then at $-1$ (becoming positive) and finally at $-4$ (becoming negative). Thus it is negative over $(-\infty,-4)$ and $(-1,2)$. Since we allow values where the expression is $0$, the final solution set is
$$
(-\infty,-4)\cup[-1,2]
$$
or, in other notation,
$$
x<-4\qquad\text{or}\qquad -1\le x\le2
$$
</s>