Rahul@gurome wrote: There are two critical points for this equation. They are x = - 5 and x = 5. Thus there are three different regions. Let's analyze each of the three regions individually,

beat_gmat_09 wrote:

Rahul@gurome wrote: There are two critical points for this equation. They are x = - 5 and x = 5. Thus there are three different regions. Let's analyze each of the three regions individually,

Hi Rahul,
Could you please explain how you came to above conclusion ?
Another doubt - you considered 3 possibilities can 4th occur ?
1) (x - 5) and (x + 5) both -ve,
2) (x - 5) -ve and (x + 5) +ve
3) (x - 5) -ve and (x + 5) both +ve
4) (x - 5) +ve and (x + 5) -ve

Thanks

goyalsau wrote:

beat_gmat_09 wrote:

Rahul@gurome wrote: There are two critical points for this equation. They are x = - 5 and x = 5. Thus there are three different regions. Let's analyze each of the three regions individually,

Hi Rahul,
Could you please explain how you came to above conclusion ?
Another doubt - you considered 3 possibilities can 4th occur ?
1) (x - 5) and (x + 5) both -ve,
2) (x - 5) -ve and (x + 5) +ve
3) (x - 5) -ve and (x + 5) both +ve
4) (x - 5) +ve and (x + 5) -ve

Thanks

I think Rahul has not considered the 4th Possibility because 2 and 4 possibility are same ,

Take this example | x |= | 2 | is given,

1) Both Positive x = 2

2) Both negative - x = -2

3) First One negative x = -2

4 ) Second one negative - x = 2

x = -2 or - x = 2 is one of the same thing.........

Rahul@gurome wrote:

beat_gmat_09 wrote:

Rahul@gurome wrote: There are two critical points for this equation. They are x = - 5 and x = 5. Thus there are three different regions. Let's analyze each of the three regions individually,

Hi Rahul,
Could you please explain how you came to above conclusion ?

Because for these two values of x, the sign of either (x - 5) or (x + 5) or both changes. On the number line, if we mark these two points we will get three different regions.

• 1. -∞ to -5
  2. -5 to 5
  3. 5 to ∞

The rest is discussed earlier.

beat_gmat_09 wrote:Another doubt - you considered 3 possibilities can 4th occur ?
1) (x - 5) and (x + 5) both -ve,
2) (x - 5) -ve and (x + 5) +ve
3) (x - 5) -ve and (x + 5) both +ve
4) (x - 5) +ve and (x + 5) -ve

Thanks

Though goyalsau tried explain, the explanation is not correct conceptually. If you observe carefully you'll find the 4th one is impossible case. If (x + 5) is negative, then (x - 5) is always positive.

Now goyalsau's explanation seems correct because if we take the 4th one as a possibility you'll find the equation boils down to the equation for 2nd one. the 4th one is not a possibility at all. So you shouldn't consider it.

@goyalsau: In the example you have given, you've done a terrible mistake. |2| = 2 always! 2 is a constant, not variable. |x| = x (for x ≥ 0) and - x (for x < 0), because we don't know the sign of x. But for 2 we know it is a positive integer. Thus |2| is always equal to 2. You cannot interpret |2| as 2 or -2.

My suggestion for absolute value problems:

• 1. Don't blindly make a list of possibilities. Most of the time this gives rise to some impossible cases. In this the impossible cases become the same as a possible case, but it may not happen always and if you are unable to identify that one you may get a incorrect result.
  2. Instead try to mark the critical points on the number line and analyze each of the region. This never gives rise to impossible cases and you will be always conceptually correct.