VITEEE 2013 Paper

|$x^2$ - x - 6| = x + 2,then Case I : $x^2$ - x - 6 < 0 ⇒ (x - 3)(x + 2) < 0 ⇒ - 2 < x < 3 In this case, the equation becomes $x^2$ - x - 6 = - x - 2 or $x^2$ - 4 = 0 ∴ x = ±2 Clearly, x = 2 satisfies the domain of the equation in this case. So, x = 2 is a solution. Case II: $x^2$ - x - 6 > 0 So, x ≤ - 2 or x > 3 In this case, the equation becomes $x^2$ - x - 6 = 0 = x + 2 i.e., $x^2$ - 2x - 8 = 0 or x = -2,4 Both these values lie in the domain of the equation in this case, so x = -2, 4 are the roots. Hence, roots are x = -2,2,4.