We have to consider two cases:
1. when x > 0 and
2. when x < 0 [Note that x cannot equal 0]
Case 1: x > 0
> 0
We isolate the radical:
>
since x > 0 we may replace |x| by x
>
The left side is non-negative and the right side is positive.
Since the left side is greater than the right side, both sides
are positive, and we can square both sides and the inequality
will not change:
- >
Add to both sides:
>
Multiply through by x² which is positive and
so the inequality will not change:
x > 2
So the only positive values of x for which the original
inequality holds is when x is greater than 2.
-------------------------------------
Case 2: x < 0
> 0
We isolate the radical:
>
Then the right side is negative and the left side
is non-negative, so the above inequality will always
be true as long as the expression under the radical
is not negative. So we only need require that the
expression under the square root radical not be
negative.
- ≥ 0
In this case |x| = -x and we have
- ≥ 0
- ≥ 0
Multiply through by x² which is positive and the
inequality will not change:
-x - 1 ≥ 0
Now we'll multiply through by -1 and the inequality
WILL change:
x + 1 ≤ 0
x ≤ -1
So the only negative values of x for which the original
inequality holds is when x is less than or equal to -1.
-------------------------
Therefore the solution in interval notation is:
(-∞,-1] ⋃ (2,∞)
Edwin
You can put this solution on YOUR website! you can solve this by graphing the equation and you can then visually see the regions that the solution lies in.
the graph looks like this.
it confirms the algebraic solution provided by edwin.
the values between -1 and 1 are not mathematically possible and therefore don't show up on the graph.