joseph0alexander wrote:
Bunuel wrote:
joseph0alexander wrote:
Bunuel,
Can we try this question by evaluating each of these expressions separately? Is this a correct approach, though time consuming?
1) x+1 > 0 ---> x > -1
2) x-1 > 0 ---> x > 1
3) -x-1 > 0 ---> -x > 1 or x > -1
If correct, could you please demonstrate on how to proceed?
This would not the best approach. You should consider two cases for |x|: when x<0, then |x| = x and when x>0, then |x| = -x. As well as two cases when (x+1) and (|x| - 1) are both positive and both negative.
Fair enough Bunuel, but just for the sake of understanding and learning the concept.
If we proceed with the 3 equations, I am getting the results in red above. Comparing them, we can take the greatest limiting factor, which is x>1, and prove sufficiency. I find this approach a little easier. Is my arithmetic correct?
Thank you.
Again you should consider 2*2 = 4 cases.
(x+1)(|x| - 1) > 0.
1. x < 0 --> (x+1)(-x - 1) > 0.(x + 1) > 0 and (-x - 1) > 0 --> x > -1 and x < -1.
No solution here.2. x < 0 --> (x+1)(-x - 1) > 0.(x + 1) < 0 and (-x - 1) < 0 --> x < -1 and x > -1.
No solution here.3. x > 0 --> (x+1)(x - 1) > 0.(x + 1) > 0 and (x - 1) > 0 --> x > -1 and x > 1.
This gives x > 1.4. x > 0 --> (x+1)(x - 1) < 0.(x + 1) < 0 and (x - 1) < 0 --> x < -1 and x < 1.
No solution here.So, (x+1)(|x| - 1) > 0 holds true only when x > 1.
Hope it's clear.