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Source: — Data Sufficiency |
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by Bidisha_800 » Mon Feb 16, 2009 10:13 am
From stmt 1

(x+1)^2 = 4(x-1)^2

(3x-1)(3-x)=0

Therefore, x =3, 1/3

INSUFF

|x-3| not = 0 INSUFF

together , we know x not equal to 3

Therefore x must be = 1/3

therefore. |x| <1


(C)
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by earth@work » Mon Feb 16, 2009 10:56 am
thanks, mod confuses me!

i did not take square of both the sides - could someone help me understand where i was going wrong in my method:

lx + 1l = 2lx - 1l
i took 2 conditions either x s negative or x is positive
when x is negative: -x+1 = 2(-x-1) ==> x= -3
when x is positive: x+1 = 2(x-1) ==> x=3
and in both the case lxl >1

i know i m making a mistake somewhere but not able to figure out what? Help needed!
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by Bidisha_800 » Mon Feb 16, 2009 11:12 am
earth@work wrote:thanks, mod confuses me!

i did not take square of both the sides - could someone help me understand where i was going wrong in my method:

lx + 1l = 2lx - 1l
i took 2 conditions either x s negative or x is positive
when x is negative: -x+1 = 2(-x-1) ==> x= -3
when x is positive: x+1 = 2(x-1) ==> x=3
and in both the case lxl >1

i know i m making a mistake somewhere but not able to figure out what? Help needed!
If you put x=-3 in the original equation you get,

|-3+1|=2|-3-1|
2=8

Therefore x=-3 is not a solution.


You didn't consider the case where 0<x<1
x+1 = -2(x-1)
x=1/3

if x >1
x+1 =2(x-1)
x=3 but according to STMT2 this is not true

so x=1/3
GMATPowerPrep Test1= 740
GMATPowerPrep Test2= 760
Kaplan Diagnostic Test= 700
Kaplan Test1=600
Kalplan Test2=670
Kalplan Test3=570
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by earth@work » Mon Feb 16, 2009 11:35 am
thanks again bidisha for help!
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