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Intern
Joined: 06 Oct 2018
Posts: 29

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30 Dec 2018, 03:09
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Hi All,

This may seem like a very stupid question, but I'll post anyways since I can't find an answer anywhere.

I have been taught to solve equations such as: x^2 = 1 as such:
x^2 = 1
x^2 - 1 = 0
(x-1)(x+1) = 0
x=1 or -1

This method works for all quadratics, and is probably the quickest way to solve them.

Why is it that when we suddenly take out the "=" and replace it with either an > or < this method falls apart?

e.g.
x^2 < 1
x^2 - 1 < 0
(x-1)(x+1) < 0

x - 1 < 0 ---> x < 1 (which is correct)
but
x + 1 < 0 ---> x < -1 (which is incorrect, should be x > -1)

solution should be: -1 < x < 1

I am aware of the wavy line method and graphical method to solve inequalities, but I'd really appreciate if someone could explain why the method outline above fails to work in these situations. I hypothesize that this method does not work because of having to consider that x could be positive or negative (much like the way absolute value equations are solved), but nonetheless, I await help from experts.

P.S., this question actually came from a DS question:

Is |x| < 1?

1) x^2 < 1
2) |x| < 1/x

obviously, in this scenario, it is easy to see that x^2 = |x| and thus statement 1 is sufficient (and so is 2 btw). But after answering the question, I went back and played around with statement 1 and then came up with this problem.

edit: I think I may have posted this in the wrong place. If so, moderators could you please move to the appropriate section?
Manager
Joined: 29 Dec 2018
Posts: 81
Location: India
WE: Marketing (Real Estate)

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30 Dec 2018, 04:08
Hi PierTotti17,

I am not an expert but can I request you to go through Bunuel 's take on this chapter?
https://gmatclub.com/forum/math-absolut ... 86462.html

_________________
Keep your eyes on the prize: 750
Math Expert
Joined: 02 Aug 2009
Posts: 7973

### Show Tags

30 Dec 2018, 05:02
PierTotti17 wrote:
Hi All,

This may seem like a very stupid question, but I'll post anyways since I can't find an answer anywhere.

I have been taught to solve equations such as: x^2 = 1 as such:
x^2 = 1
x^2 - 1 = 0
(x-1)(x+1) = 0
x=1 or -1

This method works for all quadratics, and is probably the quickest way to solve them.

Why is it that when we suddenly take out the "=" and replace it with either an > or < this method falls apart?

e.g.
x^2 < 1
x^2 - 1 < 0
(x-1)(x+1) < 0

x - 1 < 0 ---> x < 1 (which is correct)
but
x + 1 < 0 ---> x < -1 (which is incorrect, should be x > -1)

solution should be: -1 < x < 1

I am aware of the wavy line method and graphical method to solve inequalities, but I'd really appreciate if someone could explain why the method outline above fails to work in these situations. I hypothesize that this method does not work because of having to consider that x could be positive or negative (much like the way absolute value equations are solved), but nonetheless, I await help from experts.

P.S., this question actually came from a DS question:

Is |x| < 1?

1) x^2 < 1
2) |x| < 1/x

obviously, in this scenario, it is easy to see that x^2 = |x| and thus statement 1 is sufficient (and so is 2 btw). But after answering the question, I went back and played around with statement 1 and then came up with this problem.

edit: I think I may have posted this in the wrong place. If so, moderators could you please move to the appropriate section?

Hi..

Why the below doesn't work..
x - 1 < 0 ---> x < 1 (which is correct)
but
x + 1 < 0 ---> x < -1 (which is incorrect, should be x > -1)

So you have (x-1)(X+1)<0... So the product of x-1 and X+1 is less than 0.
THIS will happen if one is positive and other negative..
1) say x-1>0, then X+1<0..
x-1>0 means X>1.. and X+1<0 means X<-1..
There is no overlap in the region...we had taken X >1, so it cannot be <-1.
Hence eliminate this option.

2) x-1<0 or X<1, then X+1>0..X>-1..
So combined -1<X<1..
_________________
Intern
Joined: 06 Oct 2018
Posts: 29

### Show Tags

30 Dec 2018, 05:20
chetan2u wrote:
PierTotti17 wrote:
Hi All,

This may seem like a very stupid question, but I'll post anyways since I can't find an answer anywhere.

I have been taught to solve equations such as: x^2 = 1 as such:
x^2 = 1
x^2 - 1 = 0
(x-1)(x+1) = 0
x=1 or -1

This method works for all quadratics, and is probably the quickest way to solve them.

Why is it that when we suddenly take out the "=" and replace it with either an > or < this method falls apart?

e.g.
x^2 < 1
x^2 - 1 < 0
(x-1)(x+1) < 0

x - 1 < 0 ---> x < 1 (which is correct)
but
x + 1 < 0 ---> x < -1 (which is incorrect, should be x > -1)

solution should be: -1 < x < 1

I am aware of the wavy line method and graphical method to solve inequalities, but I'd really appreciate if someone could explain why the method outline above fails to work in these situations. I hypothesize that this method does not work because of having to consider that x could be positive or negative (much like the way absolute value equations are solved), but nonetheless, I await help from experts.

P.S., this question actually came from a DS question:

Is |x| < 1?

1) x^2 < 1
2) |x| < 1/x

obviously, in this scenario, it is easy to see that x^2 = |x| and thus statement 1 is sufficient (and so is 2 btw). But after answering the question, I went back and played around with statement 1 and then came up with this problem.

edit: I think I may have posted this in the wrong place. If so, moderators could you please move to the appropriate section?

Hi..

Why the below doesn't work..
x - 1 < 0 ---> x < 1 (which is correct)
but
x + 1 < 0 ---> x < -1 (which is incorrect, should be x > -1)

So you have (x-1)(X+1)<0... So the product of x-1 and X+1 is less than 0.
THIS will happen if one is positive and other negative..
1) say x-1>0, then X+1<0..
x-1>0 means X>1.. and X+1<0 means X<-1..
There is no overlap in the region...we had taken X >1, so it cannot be <-1.
Hence eliminate this option.

2) x-1<0 or X<1, then X+1>0..X>-1..
So combined -1<X<1..

Thank you very much Chetan. I suspected that this was the case. It is frustrating that we cannot set each expression inside the brackets to < or > than 0 like we do with regular equations.

x^2 > 1

would the approach still be valid?
i.e. the solutions to x^2 > 1 is the opposite, either x > 1 or x < -1.

x^2-1 > 0
(x-1)(x+1) > 0

hence both positive or both negative

say x-1 is negative and x+1 is also negative. Then:

x-1 < 0
x<1 (wrong)
and
x+1<0
x<-1 (right)

or both are positive

x-1>0
x>1 (right)
and
x+1>0
x>-1 (wrong)

now do we combine these sets to get the combined range?

Thanks again
Math Expert
Joined: 02 Aug 2009
Posts: 7973

### Show Tags

30 Dec 2018, 05:58
PierTotti17 wrote:
chetan2u wrote:
PierTotti17 wrote:
Hi All,

This may seem like a very stupid question, but I'll post anyways since I can't find an answer anywhere.

I have been taught to solve equations such as: x^2 = 1 as such:
x^2 = 1
x^2 - 1 = 0
(x-1)(x+1) = 0
x=1 or -1

This method works for all quadratics, and is probably the quickest way to solve them.

Why is it that when we suddenly take out the "=" and replace it with either an > or < this method falls apart?

e.g.
x^2 < 1
x^2 - 1 < 0
(x-1)(x+1) < 0

x - 1 < 0 ---> x < 1 (which is correct)
but
x + 1 < 0 ---> x < -1 (which is incorrect, should be x > -1)

solution should be: -1 < x < 1

I am aware of the wavy line method and graphical method to solve inequalities, but I'd really appreciate if someone could explain why the method outline above fails to work in these situations. I hypothesize that this method does not work because of having to consider that x could be positive or negative (much like the way absolute value equations are solved), but nonetheless, I await help from experts.

P.S., this question actually came from a DS question:

Is |x| < 1?

1) x^2 < 1
2) |x| < 1/x

obviously, in this scenario, it is easy to see that x^2 = |x| and thus statement 1 is sufficient (and so is 2 btw). But after answering the question, I went back and played around with statement 1 and then came up with this problem.

edit: I think I may have posted this in the wrong place. If so, moderators could you please move to the appropriate section?

Hi..

Why the below doesn't work..
x - 1 < 0 ---> x < 1 (which is correct)
but
x + 1 < 0 ---> x < -1 (which is incorrect, should be x > -1)

So you have (x-1)(X+1)<0... So the product of x-1 and X+1 is less than 0.
THIS will happen if one is positive and other negative..
1) say x-1>0, then X+1<0..
x-1>0 means X>1.. and X+1<0 means X<-1..
There is no overlap in the region...we had taken X >1, so it cannot be <-1.
Hence eliminate this option.

2) x-1<0 or X<1, then X+1>0..X>-1..
So combined -1<X<1..

Thank you very much Chetan. I suspected that this was the case. It is frustrating that we cannot set each expression inside the brackets to < or > than 0 like we do with regular equations.

x^2 > 1

would the approach still be valid?
i.e. the solutions to x^2 > 1 is the opposite, either x > 1 or x < -1.

x^2-1 > 0
(x-1)(x+1) > 0

hence both positive or both negative

say x-1 is negative and x+1 is also negative. Then:

x-1 < 0
x<1 (wrong)
and
x+1<0
x<-1 (right)

or both are positive

x-1>0
x>1 (right)
and
x+1>0
x>-1 (wrong)

now do we combine these sets to get the combined range?

Thanks again

Yes..
Others simply $$x^2<1... Means ....|X|<1....-1<X<1$$
And $$x^2>1...|X|>1...X<-1 .. and...X>1$$
_________________
Intern
Joined: 06 Oct 2018
Posts: 29

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30 Dec 2018, 06:08
Great, thank you for the insight and for the prompt replies Chetan!!
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