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

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what is the value of x? assume x is an integer. (1) [#permalink] New post 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...
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 [#permalink] New post 09 Sep 2005, 08:31
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.
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 [#permalink] New post 09 Sep 2005, 09:35
I think C,
Combining both and for various conditions were x could be + or - ve i think i get x could only be +2...
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 [#permalink] New post 09 Sep 2005, 09:49
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.
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 [#permalink] New post 09 Sep 2005, 10:34
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.
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 [#permalink] New post 09 Sep 2005, 11:21
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??
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 [#permalink] New post 09 Sep 2005, 12:07
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??
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 [#permalink] New post 09 Sep 2005, 13:37
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
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 [#permalink] New post 09 Sep 2005, 22:52
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 :-D :wink:

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.
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 [#permalink] New post 25 Sep 2005, 20:07
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|?
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 [#permalink] New post 26 Sep 2005, 08:01
@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
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  [#permalink] 26 Sep 2005, 08:01
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