Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 06 May 2015, 21:42

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# what is the value of x? assume x is an integer. (1)

Author Message
TAGS:
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 181 [0], given: 2

what is the value of x? assume x is an integer. (1) [#permalink]  09 Sep 2005, 07:21
what is the value of x? assume x is an integer.

(1) |x-|x^2||=2

(2) |x^2-|x||=2

pls show working...
VP
Joined: 13 Jun 2004
Posts: 1122
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 31 [0], given: 0

I would say E but i took me around 3mn

(1) |x-|x^2||=2

for each abs value you have to imagine it can be negative or positive before executing the abolute value. For example :

|x-|x^2||=2 ; x-x^2=2; x^2-x-2=0 ; (x-2)(x+1) so 2 answers : 2 and -1
x-|x^2|=2 ; no need to check this, first go to statement (2) to see if you find different numbers than in (1)

(2) |x^2-|x||=2
X^2-x=2 ; x^2-x-2=0 ; (x-2)(x+1), same answer than in statement 1

even if there are other possibilities, you've already found 2 different possible answers which are 2 and -1 in both statements so you can not choose any value for sure. E. no need to calculate everything.
Senior Manager
Joined: 29 Nov 2004
Posts: 486
Location: Chicago
Followers: 1

Kudos [?]: 9 [0], given: 0

I think C,
Combining both and for various conditions were x could be + or - ve i think i get x could only be +2...
_________________

Fear Mediocrity, Respect Ignorance

Intern
Joined: 14 Jun 2005
Posts: 37
Followers: 1

Kudos [?]: 2 [0], given: 0

Antmavel wrote:
I would say E but i took me around 3mn

(1) |x-|x^2||=2

for each abs value you have to imagine it can be negative or positive before executing the abolute value. For example :

|x-|x^2||=2 ; x-x^2=2; x^2-x-2=0 ; (x-2)(x+1) so 2 answers : 2 and -1
x-|x^2|=2 ; no need to check this, first go to statement (2) to see if you find different numbers than in (1)

(2) |x^2-|x||=2
X^2-x=2 ; x^2-x-2=0 ; (x-2)(x+1), same answer than in statement 1

even if there are other possibilities, you've already found 2 different possible answers which are 2 and -1 in both statements so you can not choose any value for sure. E. no need to calculate everything.

I dont think lx-lx^2ll=2 & lx^2-lxll=2 will have the same roots.
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 181 [0], given: 2

check you quadratic equation for (1) again.....

you have x-x^2=2

then it should follow:

x-x^2-2=0

-x^2+x-2=0

x^2-x+2=0 (you can use the quadratice equation formula)

you get x=2 or x=-1 (which you do, I am just making sure people understand what you did)

now just do what you would with (2) and you get the same factors right...

go ahead try plugging in (-1) for x in (2), it will not add upto 2!

only (2) works for x.....

you were almost there, but you have plug in the numbers to verify...

Regan used to say, "trust but verify"

Antmavel wrote:
I would say E but i took me around 3mn

(1) |x-|x^2||=2

for each abs value you have to imagine it can be negative or positive before executing the abolute value. For example :

|x-|x^2||=2 ; x-x^2=2; x^2-x-2=0 ; (x-2)(x+1) so 2 answers : 2 and -1
x-|x^2|=2 ; no need to check this, first go to statement (2) to see if you find different numbers than in (1)

(2) |x^2-|x||=2
X^2-x=2 ; x^2-x-2=0 ; (x-2)(x+1), same answer than in statement 1

even if there are other possibilities, you've already found 2 different possible answers which are 2 and -1 in both statements so you can not choose any value for sure. E. no need to calculate everything.
Intern
Joined: 27 Aug 2005
Posts: 33
Followers: 0

Kudos [?]: 0 [0], given: 0

I think the answer is C too.

For both equations 1) and 2), we get 2 answers 2 and -1.

Plugging these answers back into both the equations, we get to know that only 2 satisfies both the equations...

Whats the OA??
Current Student
Joined: 28 Dec 2004
Posts: 3391
Location: New York City
Schools: Wharton'11 HBS'12
Followers: 13

Kudos [?]: 181 [0], given: 2

dont make it sound simple....

try -2 in (2)...that also fits right?

make sure you know how to do your factorization......

nisha_qutu wrote:
I think the answer is C too.

For both equations 1) and 2), we get 2 answers 2 and -1.

Plugging these answers back into both the equations, we get to know that only 2 satisfies both the equations...

Whats the OA??
Manager
Joined: 06 Aug 2005
Posts: 197
Followers: 3

Kudos [?]: 5 [0], given: 0

It's not quite as complicated as you are making it.

(1) has roots x=2,x=-1
(2) has roots x=2,x=-2

So x=2 and C is the answer
VP
Joined: 13 Jun 2004
Posts: 1122
Location: London, UK
Schools: Tuck'08
Followers: 7

Kudos [?]: 31 [0], given: 0

Thanks for the great lesson, I was wrong. I've been too fast and assumed too easily that I got it right french people are always a little too arrogant

fresinha12 wrote:
check you quadratic equation for (1) again.....

you have x-x^2=2

then it should follow:

x-x^2-2=0

-x^2+x-2=0

x^2-x+2=0 (you can use the quadratice equation formula)

you get x=2 or x=-1 (which you do, I am just making sure people understand what you did)

now just do what you would with (2) and you get the same factors right...

go ahead try plugging in (-1) for x in (2), it will not add upto 2!

only (2) works for x.....

you were almost there, but you have plug in the numbers to verify...

Regan used to say, "trust but verify"

Antmavel wrote:
I would say E but i took me around 3mn

(1) |x-|x^2||=2

for each abs value you have to imagine it can be negative or positive before executing the abolute value. For example :

|x-|x^2||=2 ; x-x^2=2; x^2-x-2=0 ; (x-2)(x+1) so 2 answers : 2 and -1
x-|x^2|=2 ; no need to check this, first go to statement (2) to see if you find different numbers than in (1)

(2) |x^2-|x||=2
X^2-x=2 ; x^2-x-2=0 ; (x-2)(x+1), same answer than in statement 1

even if there are other possibilities, you've already found 2 different possible answers which are 2 and -1 in both statements so you can not choose any value for sure. E. no need to calculate everything.
Intern
Joined: 25 Jun 2005
Posts: 23
Location: Bay Area, CA
Followers: 0

Kudos [?]: 0 [0], given: 0

Antmavel wrote:
(2) |x^2-|x||=2
X^2-x=2 ; x^2-x-2=0 ; (x-2)(x+1), same answer than in statement 1

How do you convert |x^2-|x||=2 to
x2 - x = 2 ?
ignore abs completely?
Dont we have to do
x2 - |x| = 2
and x2 - |x| = -2
first, and then 2 more equations for the |x|?
_________________

If you can't change the people, change the people.

Intern
Joined: 05 Sep 2005
Posts: 11
Followers: 0

Kudos [?]: 0 [0], given: 0

@Dil66 and all:

(2) |x^2-|x||=2

There are 2 cases:
(a) x>=0: Then we have |x^2-x|=2
(a1) x^2-x=-2 (no root)
(a2) x^2-x=2 (roots are 2 (-1 is not since x>=0))

(b) x<0: Then we have |x^2+x| =2
(b1) x^2+x=-2 (no root)
(b2) x^2 +x=2 (roots are -2 (1 is not since x<0))

(2) |x-|x^2||=2

<=> |x-x^2|=2 (since x^2 is always greater or equal 0)
(a) x-x^2=-2 (roots are 2 and -1)
(b) x-x^2 = 2 (no root)

So the answer is C with x=2

Hope this clear
_________________

I can, therefore I am

Similar topics Replies Last post
Similar
Topics:
2 If x > 1, what is the value of integer x? (1) There are x 6 01 Jul 2011, 11:39
If x is an integer, what is the value of x? (1) 3 02 May 2011, 04:00
5 If x > 1, what is the value of integer x? 7 18 Mar 2011, 13:07
3 If x>1, what is the value of integer x? 15 10 Mar 2011, 05:50
If x is an integer, what is the value of x? (1) 12 25 Apr 2008, 18:53
Display posts from previous: Sort by