Question

Write solution set of |x + (1/x)| > 2 ; x belongs to real numbers
Please help me with this question.
thank you.

Answer 1

We have |x+$\frac{1}{x}$| > 2. Combining the terms we get
|$\frac{x^2+1}{x}$| >2.
Note that the numerator is always positive due to the square. hence we can write
$\frac{x^2+1}{|x|}$ > 2. We can multiply both sides by |x| and it is positive so does not change the inequality:
$x^2+1$ > 2|x| (1)
Let us consider 2 case: x>0 and x <0
Case 1: x >0
Here equation (1) becomes $x^2$+1 >2x
$x^2$-2x+1 > 0
$(x-1{)}^2$ > 0
But because of the square, this is always true except when x-1= 0 or x = 1
So for x>0, the solution is all x > 0 except x = 1.
Case 2: x <0
Here equation (1) becomes $x^2$+1 >-2x since for x < 0, |x| = -x
So $x^2$+2x+1 > 0
$(x+1{)}^2$ > 0
Again, because of the square, this is always positive except when x +1=0 or x = -1
So for x<0, the solution is all x <0 except x = -1.
Finally, we hae to exclude the point x = 0, since 1/x would be undefined at that point.
The complete solution is therefore:
x can be any real number except x= -1,x=0 and x=1