The Set $x:\left |x+\frac{1}{x}\right|>6=?$

Question

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?

Answer 1

Hint $$|f(x)|>A \Leftrightarrow$$ $f(x)>A$ or $f(x)<-A$

Or so:

$x>0 \Rightarrow \left(\frac{x^2+1}x-6 \right)>0$

$x<0 \Rightarrow \left(-\frac{x^2+1}x-6 \right)>0$

Answer 2

Hint: $$|f(x)|>A \iff f(x)>A \text{ or } f(x) < -A \iff (f(x))^2>A^2$$ when $A$ is positive.

Details:

I think this will be an easier method to solve since you don't need to take cases and solve twice. Square the given equation, multiply throughout by $x^2$ and substitute $x^2=t$ to make it a quadratic. Solve the quadratic inequality by factoring the quadratic and then put $t=x^2$ back and solve.

Feel free to comment if you have any problems with this.