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

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
$$x\le-\frac43.$$
