GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Oct 2019, 20:56 GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Intern  B
Joined: 06 Oct 2018
Posts: 29

Show Tags

1 00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

HideShow timer Statistics

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  S
Joined: 29 Dec 2018
Posts: 81
Location: India
WE: Marketing (Real Estate)

Show Tags

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

Also, this article https://gmatclub.com/forum/absolute-mod ... l#p1622372 by chetan2u might also interest you. (<---- critical method here might help you)
_________________
Keep your eyes on the prize: 750
Math Expert V
Joined: 02 Aug 2009
Posts: 7973

Show Tags

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  B
Joined: 06 Oct 2018
Posts: 29

Show Tags

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.
Had the question been:

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 V
Joined: 02 Aug 2009
Posts: 7973

Show Tags

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.
Had the question been:

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  B
Joined: 06 Oct 2018
Posts: 29

Show Tags

Great, thank you for the insight and for the prompt replies Chetan!! Display posts from previous: Sort by

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne  