x 2 greaterthanoreqaualto x 2 2 6SolutionWehave x 2 x2

|x + 2| greaterthanoreqaualto | x + 2 |^2 - 6

Solution

Wehave | x +2| | x+2|² -6.

Let | x +2| = a. Then the equation chages to a a²-6 or, a²–a -6 0 or, a² –3a +2a -6 0 or, a(a-3)+2(a-3a) 0 or, (a-3)(a+2) 0 . Now,there are 2 possibilities:

However, since a = | x +2| 0, hence a+2 0. Thus the only possibility is (a-3) 0 and (a+2) 0. Thus, a 3 and a -2 i.e. -2 | x +2| 3

If | x +2| 3, then -3 (x+2) 3 so that -5 x 1

If | x +2| -2 , then -2 (x+2) 2 or, -4 x 4 .

Thus, if -2 | x +2| 3, then -4 x 1 or x [ -4,1]