$x^2 - | x-2 | + 6 > 0 $ , where x belongs to $R$
I am not sure about my own approach to this ques. I solved it as:
$x^2 + 6 > | x-2 |$ , thereafter i got 2 cases
Case 1: $-(x^2 + 6) < (x - 2)$ , which finally gave, $x^2 + x + 4 > 0 $
Case 2: $ (x^2 + 6) > (x - 2)$ , which finally gave, $x^2 - x + 8 > 0 $
Both the above equations have no Real solutions, and since they have asked the Range of x and not $F(x)$, is it correct to say : Range of x is a particular set of Complex Nos. with an empty set of Real Nos.
All suggestions are welcome.
Thanks.
Both your inequalities are satisfied by any real number, since they are quadratic graphs that lie over the $x$-axis.
All in all, your inequality is satisfied over the whole of $\mathbb{R}$, a sketch should be enough to convince you. (blue is $x^2 +6$ and $|x-2|$ is in red)