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!
 A: Hint : You wrote $\frac{x^3+4x^2-2x-8-x^3-2x}{(x^2+2)(x+4)} \geq 0 \Leftrightarrow$ $4x^2-4x-8 \geq 0$ but you forgot to study what is on the bottom. $(x+4)$ changes sign.
A: 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,
$$
x<-4\qquad\text{or}\qquad -1\le x\le2
$$
A: It might be useful to simplify the inequality before trying to solve:
$$\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4} \Leftrightarrow x^2-2 \leq \frac{x^3+2x}{x+4}$$
Now, note that (using polynomial division)


*

*$\frac{x^3+2x}{x+4} = x^2 - 4x+18 -\frac{72}{x+4}$


So, the inequality turns into 
$$x^2-2 \leq x^2 - 4x+18 -\frac{72}{x+4} \Leftrightarrow \frac{72}{x+4} \leq -4x+20 = 4(5-x) \Leftrightarrow \boxed{\frac{18}{x+4} \leq 5-x}$$
Taking into consideration that the inequality sign flips when multiplying by a negative number you get two cases:
$$\frac{18}{x+4} \leq 5-x \Leftrightarrow \begin{cases} 18 \leq (5-x)(x+4) & \mbox{for } x >-4 \\ 18 \geq (5-x)(x+4) & \mbox{for }x<-4\end{cases} \Leftrightarrow \boxed{\begin{cases} (x+1)(x-2) \leq 0 & \mbox{for } x >-4 \\ (x+1)(x-2) \geq 0 & \mbox{for }x<-4\end{cases}}$$
Note that $(x+1)(x-2)$ corresponds to an upwards open parabola with zeros at $x=-1$ and $x=2$. So, you get
$$\begin{cases} (x+1)(x-2) \leq 0 & \mbox{for } x >-4 \\ (x+1)(x-2) \geq 0 & \mbox{for }x<-4\end{cases} \Leftrightarrow \begin{cases} x \in [-1,2] \cap (-4, \infty)\\ x \in ((-\infty,-1] \cup [2,\infty)) \cap ( -\infty,-4) \end{cases} \Leftrightarrow \boxed{x\in ([-1,2] \cup ( -\infty,-4))}$$
A: You obtained the equivalent form $$\frac{4x^2-4x-8}{(x^2+2)(x+4)}\geq 0$$
Here, $x^2+2$ in the denominator is positive for all $x\in \mathbb{R}$, so you can cancel it. 
However, you need to make a case distinction in terms of the sign of $x+4$.
A: You reduced the inequality to $$\frac{4(x+1)(x-2)}{(x^2+2)(x+4)} \ge 0$$
which is equivalent to $$\frac{(x+1)(x-2)}{x+4} \ge 0$$
This is true if and only if an even number of terms $(x+1), (x-2), (x+4)$ are $\le 0$. There are four options:


*

*$x+1 \ge 0$, $x-2 \ge 0$, $x+4 > 0$ which gives $x \in [2, +\infty\rangle$

*$x+1 \le 0$, $x-2 \le 0$, $x+4 > 0$ which gives $x \in \langle -4, -1]$

*$x+1 \le 0$, $x-2 \ge 0$, $x+4 < 0$ which gives no solutions

*$x+1 \ge 0$, $x-2 \le 0$, $x+4 < 0$ which gives no solutions


Therefore the solutions are $x \in \langle -4, -1] \cup [2, +\infty\rangle$.
