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

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Current Student
Joined: 28 Dec 2004
Posts: 3357
Location: New York City
Schools: Wharton'11 HBS'12
what is the value of x? assume x is an integer. (1) [#permalink]

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09 Sep 2005, 08:21
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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: 1115
Location: London, UK
Schools: Tuck'08

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09 Sep 2005, 09:31
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: 483
Location: Chicago

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09 Sep 2005, 10:35
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

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09 Sep 2005, 10:49
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: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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09 Sep 2005, 11:34
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

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09 Sep 2005, 12:21
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: 3357
Location: New York City
Schools: Wharton'11 HBS'12

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09 Sep 2005, 13:07
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

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09 Sep 2005, 14:37
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: 1115
Location: London, UK
Schools: Tuck'08

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09 Sep 2005, 23:52
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

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25 Sep 2005, 21:07
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

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26 Sep 2005, 09:01
@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

26 Sep 2005, 09:01
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