<s>[INST] Domain of inequation inside squared root How should I go when restricting the roots of the inequation:
$\sqrt {x^2+5x+6} - \sqrt {x^2-x+1} \lt 1$?
By restricting both the squared roots, I know that:
$x \le 3$ and $x \ge -2$
However when simplifying the whole inequation, I get the two roots:
$\frac{-13-\sqrt{73}}{16}$ and $\frac{-13+\sqrt{73}}{16}$.
Both roots are valid when swapping them in the first inequation, so how should I restrict my $x$? Should the final answer be: $x \le 3$ and $x \ge -2$?
 [/INST] It's $$\sqrt{x^2+5x+6}<1+\sqrt{x^2-x+1}$$ or
$$x^2+5x+6<1+x^2-x+1+2\sqrt{x^2-x+1}$$ or
$$\sqrt{x^2-x+1}>3x+2.$$
Now, consider two cases:


*

*$x< -\frac{2}{3},$ which is $x\leq-3$ or $-2\leq x<-\frac{2}{3}$;

*$x\geq-\frac{2}{3}.$
Can you end it now?
I got the following answer:
$$(-\infty,-3]\cup\left[-2,\frac{-13+\sqrt{73}}{16}\right]$$
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