Solve the following inequality:
$$\displaystyle \sqrt{x \, + \, 61} \, < \, x \, + \, 5.$$

Solution

$$x+61\geq0$$
$$x\epsilon[-61,\infty)$$
$$\sqrt{x+61}<x+5$$
Squarring both sides
$$x+61<x^2+10x+25$$
$$x^2+9x-36>0$$
$$(x+12)(x-3)>0$$
$$x\epsilon (-\infty,-12)\cup(3,\infty)$$
$$x\epsilon[-61,-12)\cup(3,\infty)$$