2025/023) Solve $\sqrt{(x+2)} > x $

It is risky to square both sides as it shall give erroneous root.

Let us first find the range of x

as we can take the square root of a non -ve number only so we 

$x \ge -2$ 

for $x \in [-2,0] LHS is positve and RHS is -ve or zero. so  this is range in  -ve

To find the range is positive we square both sides to get 

$(x+2) > x^2$

or   $x^2-x-2 <0$

or $(x-2)(x+1) <0$ or   $x \in [-1,2)$

as x is positive so   $x \in [0,2)$

so combining we get  $x \in [-2,2)$