The Set $x:\left |x+\frac{1}{x}\right|>6=?$ The question is that ,the Set $x:\left |x+\frac{1}{x}\right|>6$ equals what intervals of $x$?
My approach:-
I tried to solve the inequality and get interval for $x$'s value as follows:-
$$\left|\frac{x^2+1}{x}\right|-6>0$$
case 1.)
if $x$ is -ive-
$$-x^2-1-6x>0\implies x^2+6x+1<0\implies(x-(-3+2\sqrt2))(x-(-3-2\sqrt2))<0$$so that interval is 
$$(-3-2\sqrt2,-3+2\sqrt2)\tag{1.}$$
case2.)
if $x$ is +ive$$x^2-6x+1>0\implies(x-(3-2\sqrt2))(x-(3+2\sqrt2))>0$$
so that the interval is -:
$$(0,3-2\sqrt2)\cup(3+2\sqrt2,\infty)\tag{2.}$$
The answer would be union of interval (1.) &(2.) but this answer is not in options.
So what is the correct solution of this question? What would be the interval?

Hint $$|f(x)|>A \Leftrightarrow$$
$f(x)>A$ or $f(x)<-A$
Or so:
$x>0 \Rightarrow \left(\frac{x^2+1}x-6 \right)>0$
$x<0 \Rightarrow \left(-\frac{x^2+1}x-6 \right)>0$

