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 $x \ge -2$?
 A: 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:
$$(-\infty,-3]\cup\left[-2,\frac{-13+\sqrt{73}}{16}\right]$$
A: As $x^2-x+1 > 0$ it has no effect on the domain. However, $x^2+5x+6$ is negative in the interval $(-3,-2)$. So we must keep this in mind when we are crafting a solution
Consider $$\sqrt{(x+2)(x+3)} < 1+\sqrt{x^2-x+1}$$
$$x^2+5x+6 < x^2-x+2 + 2\sqrt{x^2-x+1}$$
$$3x+2 < \sqrt{x^2-x+1}$$
$3x+2$ being linear and $\sqrt{x^2-x+1}$ being a function with only one critical point there are at most two points where these expressions are equal (note that both their squares are quadratic)
Consider the points where
$$9x^2 + 4 + 12x =x^2 -x+1$$
$$\implies 8x^2 + 13x +3 = 0$$
$$\implies x = {-13\pm\sqrt{73}\over16}$$
As you have found. 
(Note that $\sqrt{73}$ is between 8 and 9)
So, everywhere that is not between these points and is in the domain is in our solution. Again using our estimate for the roots, this gives:
$$\boxed{x\in (-\infty,-3]\cup\left[-2,{-13+\sqrt{73}\over 16}\right)}$$
A: Break the 2 expressions within roots into linear factors and decide when both the expressions in the roots are simultaneously nonnegative.
