<s>[INST] Find the domain of $\ln(\sqrt{x^2-5x-24}-x-2)$. Find the domain of $\ln(\sqrt{x^2-5x-24}-x-2)$.
When i solved this question,i got the answer $x<\frac{-28}{9}$ but the answer given in the book is $x\leq-3$.
This is how i solved.
$\sqrt{x^2-5x-24}-x-2>0\Rightarrow \sqrt{x^2-5x-24}>x+2$
$$\therefore x^2-5x-24>x^2+4x+4\Rightarrow9x<-28\Rightarrow x<\frac{-28}{9} \tag 1$$
and
$$x^2-5x-24\geq0\Rightarrow x\leq -3 \text{ or } x\geq 8 \tag 2$$
When taking the intersection of (1) and (2),we get $x<\frac{-28}{9}$. Where have i gone wrong? Please help me.
 [/INST] First note $\sqrt{x^2-5x-24}$ is a real number iff $x\in(-\infty, -3]\cup[8,\infty)$
Notice that 
$$x\ge 8\,\,\text{ and }\,\,\sqrt{x^2-5x-24}>x+2\quad \implies\quad x^2-5x-24>(x+2)^2\iff 9x+28<0$$
which is imposible.
On the other hand
$$x\le -3 \implies\quad \sqrt{x^2-5x-24}\ge 0 \quad\text{ and }\,\,\,-1 = -3 +2 \ge x+2$$
Then, $$\sqrt{x^2-5x-24}-x-2\ge 1>0$$
So, $\ln(\sqrt{x^2-5x-24}-x-2)$ is a real number iff $x\le -3$.
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