Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$ 
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stuck here.
 A: Look at your original equation and original solution of $x \leq -4$: 


*

*what happens when $x = -4$? The fraction is not defined there. 

*What about $x = 4?$ That satisfies the original equation. But it is not accounted for in your solution.


Think about why your solution is (was: since edited) problematic, and where you went astray. Try multiplying by the factor of $(x + 4)$: Note when $x\lt 4$, the factor is negative (i.e.) multiplying by $(x + 4)\lt 0$ reverses the direction of the inequality. When $x \gt -4$, then $(x + 4) > gt 0$, and so the direction of the inequality remains unchanged.
So multiply by $(x + 4)$, but consider both possibilities:


*

*Solve for $x$ when $$(x + 4) > 0,\; \iff x \gt -4$$ 

*Solve for $x$ when $$(x+4) \lt 0 \iff x < -4,$$


You should find that the inequality is true/satisfied whenever $x \lt -4, \text{ or}\;\; x \geq -2$
And remember: we cannot have $x = -4!$.
A: No. You should have $2(x+4)^2$ on the right after multiplying. Then $$0\le2(x+4)^2-4(x+4)\\0\le(x+4)^2-2(x+4)\\0\le(x+4)\bigl((x+4)-2\bigr)\\0\le(x+4)(x+2).$$ Now, that last inequality is true whenever $x+4$ and $x+2$ have the same sign or one of them is $0$--that is, whenever $x+4\le0$ ($x\le-4$) or $x+2\ge0$ ($x\ge -2$)--but we can't allow $x=-4$ in our original inequality. Hence, our given inequality is true whenever $x<-4$ or $x\ge -2$.
A: $$ \frac{4}{x+4} < 2 $$
(We'll do a case-by-case analysis here, which I feel is slightly more illuminating.)
Case 1: $x+4 > 0 \quad$ ($x > -4$)
In this case, we just multiply both sides by $x+4$ to get:
$$ 4 \le 2(x+4) $$
$$ 4 \le 2x + 8 $$
$$ -4 \le 2x $$
$$ -2 \le x $$
So this solution occurs when both $-4 < x$ and $-2 \le x$, which is the same condition as $x \ge -2$. (Draw a number line to see why this is the case.)
Case 2: $x+4 < 0 \quad$ ($x < -4$)
Now, we must flip the inequality when we multiply both sides.
$$ 4 \le 2(x+4) $$
$$ 4 \le 2x + 4 $$
$$ 0 \le 2x $$
$$ 0 \le x $$
$$ x \ge 0 $$
We have a solution whenever $x < -4$ and $x \ge 0$, in other words, $x < -4$. 
A: An idea:
$$\frac{4}{x+4}\le 2\stackrel{\text{div. by 2}}\iff \frac{2}{x+4}\le 1\iff \frac{2}{x+4}-1\le 0\stackrel{\text{common denom.}}\iff $$
$$\iff \frac{2-x-4}{x+4}\le 0\stackrel{\text{mult. by (-1)}}\iff \frac{x+2}{x+4}\ge0\stackrel{\text{mult. by}\; (x+4)^2}\iff(x+2)(x+4)\ge 0\iff$$
$$\iff x<-4\,\,\vee\,\,x\ge-2\;\ldots$$ 
A: Multiplying by $(x+4)^2$ solves the worry about whether it will be negative (though you will still have problems at $x=-4$, not a worry here).  You can now subtract $4x$ from both sides.  But I would multiply by $x+4$.  You need to split the cases, as when $x \lt -4$ this is negative and you need to reverse the inequality.  So $$\frac 4{x+4}\le 2$$  Assume $x \lt -4$then $ 4 \ge 8+2x$ and this must be true, alternately let $x \gt -4$ and we have $4 \le 8 +2x$ or $-2 \le x$.  So $x \lt -4$ or $-2 \le x$ is the final answer.
A: Hint: Complete the square in your last expression $2x^2 + 12x + 16$.
Solution: 
Completing the square gives 
$$2x^2 + 12x + 16 = 2(x^2 + 6x + 4) = 2(x^2 + 2\cdot 3x + 3^2 - 1) = 2(x+3)^2 - 2$$
You already showed that your inequality is equivalent to $2x^2 + 12x + 16 \geq 0$.
Using the result of completing the sqare, we continue: $$2x^2 + 12x + 16 \geq 0 \\ \iff (x+3)^2 - 1 \geq 0 \\ \iff (x+3)^2 \geq 1 \\ \iff \left| x+3\right| \geq 1 \\ \iff \left| x - (-3)\right| \geq 1 \\ \iff \text{the distance between }x\text{ and }-3\text{ is at least }1 \\\iff x \leq -4 \text{ or } x\geq -2.$$
By the fact that $x = -4$ is forbidden, we can also write
$$x < -4 \text{ or }x\geq -2.$$
Comment:
I like your start where you multiply by $(x + 4)^2$ (and not just by $x+4$). In this way, you avoid the case-by-case analysis!
A: Why not just do it this way:
For $x + 4 > 0$ the following is legal:
$$
\begin{align*}
\frac{4}{x + 4} &\le 2 \\
\frac{x + 4}{4} &\ge \frac{1}{2} \\
x + 4 &\ge 2 \\
x &\ge -2
\end{align*}
$$
If $x + 4 < 0$, i.e. $x < -4$, the inequality is satisfied.
The solution is $(-\infty, -4) \cup [-2, \infty)$.
A: Just looking at $\dfrac{4}{x+4}$, we can see that if $x\lt -4$, then $\dfrac{4}{x+4}$ will be negative, and therefore $\le 2$.
Now suppose that $x\gt -4$. Then the components are pleasantly positive.  We can safely multiply through by $x+4$, getting $4\le 2x+8$, or equivalently $x\ge -2$. 
Remark: Multiplying by $(x+4)^2$ is, apart from the easily dealt with issue at $x=-4$, a technically valid step. However, it leads to a mildly complicated expression, with manifold possibilities of error.
That is not my primary objection. What concerns me is operating mechanically, giving oneself over to the algebra. It is all too easy to lose contact with the ground.
Doing things semi-mechanically can be a useful strategy for end-of-section problems. However, it is not a viable strategy if one really wants to use mathematics. 
