Solving $\frac{x^2-2}{x^2+2} \leq \frac{x}{x+4}$.

Question

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!

Answer 1

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, $$ x<-4\qquad\text{or}\qquad -1\le x\le2 $$

Answer 2

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.

Answer 3

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 $x+4$.

Answer 4

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

Answer 5

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

$$\begin{cases} (x+1)(x-2) \leq 0 & \mbox{for } x >-4 \\ (x+1)(x-2) \geq 0 & \mbox{for }x<-4\end{cases} \Leftrightarrow \begin{cases} x \in [-1,2] \cap (-4, \infty)\\ x \in ((-\infty,-1] \cup [2,\infty)) \cap ( -\infty,-4) \end{cases} \Leftrightarrow \boxed{x\in ([-1,2] \cup ( -\infty,-4))}$$