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Inequalities - Is X^2 > 5^2

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by Rahul@gurome » Mon Dec 06, 2010 6:56 am
pradeepspanchal wrote:Is X^2 > 5^2 ?
1) |X -5| = 3 |X + 5|
2) |X| > 3
Question is asking whether x² > 5² = 25 or not. Which happens when x < -5 or x > 5.

Statement 1: |x - 5| = 3|x + 5|
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,
  • 1. x < -5
    • (x - 5) and (x + 5) both are negative, thus
      |x - 5| = 3|x + 5| => -(x - 5) = -3(x + 5) => 2x = -20 => x = -10
    2. -5 ≤ x < 5
    • (x - 5) is negative but (x + 5) is positive, thus
      |x - 5| = 3|x + 5| => -(x - 5) = 3(x + 5) => 4x = -10 => x = -10/4 = -2.5
    1. x ≥ 5
    • (x - 5) and (x + 5) both are positive, thus
      |x - 5| = 3|x + 5| => (x - 5) = 3(x + 5) => 2x = -20 => x = -10 (Not possible as we assumed x ≥ 5)
Thus only two possible values of x.
For x = -10, x² > 25
But for x = -2.5, x² < 25

Not sufficient.

Statement 2: |x| > 3
Implies x > 3 or x < -3.
Thus x² may or may not be greater than 25.

Not sufficient.

1 & 2 Together: As |x| > 3, only possible value of x is -10 => x² > 25

Sufficient.

The correct answer is C.
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by pradeepspanchal » Mon Dec 06, 2010 8:19 am
Thanks Rahul !!!
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by beat_gmat_09 » Mon Dec 06, 2010 9:39 am
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
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by goyalsau » Mon Dec 06, 2010 10:23 am
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.........
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by beat_gmat_09 » Mon Dec 06, 2010 8:38 pm
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.........
Thanks ! It helps.
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by Rahul@gurome » Mon Dec 06, 2010 10:53 pm
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.
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by goyalsau » Mon Dec 06, 2010 11:26 pm
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.
Thanks Rahul, Can you please Give us a example How to use Number line on this one????
Saurabh Goyal
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by Rahul@gurome » Mon Dec 06, 2010 11:41 pm
goyalsau wrote:Thanks Rahul, Can you please Give us a example How to use Number line on this one????
For this particular problem as I have discussed mark -5 and 5 on number line. And analyze the three regions (as mentioned in the earlier post).

For your example, |x| = |2|, we can simply write it as |x| = 2, as |2| is always equal to 2. This is now a very simple problem, there is no need to introduce number line here. We can introduce and analyze the problem with number line, but that will be an unnecessary complication. The concept of number line is useful when there are more than one critical points, for example this particular problem.

If you insist to know how number line can be used in your example, follow me!
  • 1. The equation has one critical point, x = 0 (Where the expression within the modulus changes sign)
    2. We have two region x < 0 and x ≥ 0.
    3. For x < 0, the equation becomes -x = 2 => x = -2
    4. For x ≥ 0, the equation becomes x = 2
    5. Two solutions x = -2 and x =2
As I said for this simple problem there is no need to bring such complication!
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by beat_gmat_09 » Tue Dec 07, 2010 4:52 am
Thanks Rahul, for detailed explanation.
Helpful.
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