$\sqrt{9x^2-16}>3x+1$ I'm trying to solve the following inequality:

$$\sqrt{9x^2-16}>3x+1$$


Here's my attempt:
$\sqrt{9x^2-16}>3x+1$
$\longrightarrow 9x^2-16>9x^2+6x+1$
$\longrightarrow -16>6x+1$
$\longrightarrow x<-\frac{17}{6}$
Now, I need to check the constraints:
$9x^2-16 > 0$
$\longrightarrow (3x)^2 > 4^2$
$\longrightarrow \pm3x > 4$
$\longrightarrow 3x > 4$; $-3x > 4$
$\longrightarrow x > \frac{4}{3}$; $x < -\frac{4}{3}$
Making sure the answer meets the constraints:
$\{(-\infty, -\frac{4}{3})\cup(\frac{4}{3}, \infty)\}\cap (-\infty, -\frac{17}{6}) = (-\infty, -\frac{17}{6})$
So, my answer is $x=(-\infty, -\frac{17}{6})$, however verifying on Wolfram|Alpha results in $x=(-\infty, -\frac{4}{3}]$.

Where, what, and why is wrong with my solution?
 A: First the domain of the inequation is $(-\infty,-4/3]\cup[4/3,+\infty)$.
Next, you should know that, when $A\ge 0$,
$$\sqrt A>B\iff A >B^2\quad\text{or}\quad B<0.$$
In the present case, one gets
\begin{cases}
9x^2-16>(3x+1)^2\iff x<-\dfrac{17}6 \\
\quad \text{or}\\
x<-\dfrac13.
\end{cases}
Both conditions yield $x\in (-\infty,-1/3)$. Taking the domain of validity of the inequation into account, one obtains
$$x\le-\frac43.$$
A: *

*$3x+1\leq0$ and $9x^2-16\geq0$, which gives $x\leq-\frac{4}{3}$;

*$3x+1>0$ and $9x^2-16>(3x+1)^2$. The last gives $x<-\frac{17}{6}$, which is impossible. 
Thus, the answer is $\left(-\infty,-\frac{4}{3}\right]$.
A: at first it must be $|x|\geq \frac{4}{3}$ if $$x\le -\frac{1}{3}$$ our inequality is true. let $$x>-\frac{1}{3}$$ we can square it:
$$9x^2-16>9x^2+6x+1$$ otr $$x<-\frac{17}{6}$$ together with the conditions $$-\infty<x<-\frac{17}{6}$$ and $$-\frac{1}{3}<x$$ or $$-\frac{1}{3}\geq x$$ and $$x\geq \frac{4}{3}$$ or $$x\le -\frac{4}{3}$$ we get $$-\infty<x\le- \frac{4}{3}$$ as the searched solution set
A: $a > b \implies a^2 > b$ is not true if $b < 0$.
We have two cases: 
1) if $3x + 1 \ge 0$ or $x \ge -\frac 13$ you correctly got $x < \frac {-17}6$.  But that contradicts $x \ge -\frac 13$
So we must have case 2:
2)  $3x + 1 < 0$ and $x < -\frac 13$.  
From that we can't square both sides as $a \ge 0 > b$ does/can not tell us anything about $a^2$ in relation to $b^2$ as $|a|$ and $|b|$ can be any relation.
In other words we have no idea if $\sqrt{x^2 - 16}^2 >,<, = (3x + 1)^2$.  Any one of those is possible.
But we do have $9x^2 - 16 \ge 0$ and $x < -1/3$.  That will get you $x^2 \ge \frac {16}9$ so $|x| \ge \frac 43$ so either $x \ge \frac 43$ which is is a contradiction, or $x \le -\frac 43$.  So $x \le -\frac 43$
So we have $x \le -\frac 43$ and $x < -\frac 13$.  So solution is $x \le -\frac 43$.
