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Mod problem - anything wrong with this approach

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by pbanavara » Mon Nov 24, 2008 6:51 pm
is |x| < 1 ?

|x+1| = 2|x-1|
|x-3| != 0

Taking option 1:

x+1 = 2(x-1) or x+1 = 2(1-x) or -x-1 = 2(x-1) or -x-1 = 2(1-x)

Solving x=1/3 or x=3 Not sufficient

Option be x-3 !=0 or 3-x!=0 => x != 3 not sufficeint

Combining both x=3 which answers the question .. but the OA is different - what could be wrong here ..

Thanks,
Pradeep
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by cramya » Mon Nov 24, 2008 6:56 pm
Is the OA E)?
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by pbanavara » Mon Nov 24, 2008 7:02 pm
cramya wrote:Is the OA E)?
Absolutely
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by cramya » Mon Nov 24, 2008 7:04 pm
I hope I am not missing something here.

Stmt II

|x-3| !=0 means x is not 3. It can be anything else to satisfy this equation

Stmt I

Like u said gives 2 values for x

Combining them leads us to a world of possibilities

INSUFF

E)
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by pbanavara » Mon Nov 24, 2008 7:16 pm
cramya wrote:I hope I am not missing something here.

Stmt II

|x-3| !=0 means x is not 3. It can be anything else to satisfy this equation

Stmt I

Like u said gives 2 values for x

Combining them leads us to a world of possibilities

INSUFF

E)
Sorry I mean't : from option1 x = 3 or 1/3 from option 2 x !=3 So doesn't it imply x= 1/3 ????
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by cramya » Mon Nov 24, 2008 7:23 pm
I see where u r coming form now. I will try to think some more and hopefully I can come up wiht an answer.

My thought was from stmt II x could be, to name a few 4,5,6 etc..

Combining we know x is not 3 but it could be 1/3,4,5,6,7 etc..

I am hoping others will respond too...
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by pbanavara » Mon Nov 24, 2008 7:31 pm
cramya wrote:I see where u r coming form now. I will try to think some more and hopefully I can come up wiht an answer.

My thought was from stmt II x could be, to name a few 4,5,6 etc..

Combining we know x is not 3 but it could be 1/3,4,5,6,7 etc..

I am hoping others will respond too...
The more I think .... it's something like :

x = a, x=b
x!=b - does not necessarily mean x=a x could be c .. which I think is more in line with your explanation.

Boy this is really tricky ..
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Re: Mod problem - anything wrong with this approach

by vittalgmat » Mon Nov 24, 2008 9:25 pm
pbanavara wrote:is |x| < 1 ?

|x+1| = 2|x-1|
|x-3| != 0

Taking option 1:

x+1 = 2(x-1) or x+1 = 2(1-x) or -x-1 = 2(x-1) or -x-1 = 2(1-x)

Solving x=1/3 or x=3 Not sufficient

Option be x-3 !=0 or 3-x!=0 => x != 3 not sufficeint

Combining both x=3 which answers the question .. but the OA is different - what could be wrong here ..

Thanks,
Pradeep

Here is something I learnt from this forum:
if u have statement with 2 abs on either sides of =, then u can find the boundary values by equating each side to 0 and solving. (ie. x +1 = 0 -> x = -1;
2 (x -1) = 0
2x -2 = 0
2x = 2
x =1.
Now on a number line mark off the two values of x. U will get 3 regions.
Solve for values of x for each region.



I have another question
can |x -1| = 2 |x +1| be squared on both sides to get rid of the ABS symbols?? I remember seeing something like this in the forum.
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by vittalgmat » Mon Nov 24, 2008 9:34 pm
OOPS!!!
the method that I suggested is applicable for inequalities
ie for |x +1| < 2|x -1|

thanks
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by pbanavara » Mon Nov 24, 2008 10:28 pm
vittalgmat wrote:OOPS!!!
the method that I suggested is applicable for inequalities
ie for |x +1| < 2|x -1|

thanks
Yep I was about to say so :)
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by vittalgmat » Mon Nov 24, 2008 10:32 pm
Any thoughts on this question ???


can |x -1| = 2 |x +1| be squared on both sides to get rid of the ABS symbols?? I remember seeing something like this in the forum.

Thanks
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by Tryingmybest » Wed Nov 26, 2008 12:44 pm
vittalgmat - try doing it , you will get imaginary roots , So insufficient
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by austin » Wed Nov 26, 2008 8:29 pm
Tryingmybest,

Statement 1: |x+1| = 2|x-1|
Squaring, (x+1)^2 = 4(x-1)^2
=>x^2+1+2x = 4x^2+4-8x
=> 3x^2-10x+3 = 0
=> 3x^2-9x-x+3 = 0
=> 3x(x-3) -1(x-3) = 0
=> (3x-1) (x-3) = 0

How do you get img. roots?
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by cramya » Thu Nov 27, 2008 3:47 pm
Any thoughts on this question ???


can |x -1| = 2 |x +1| be squared on both sides to get rid of the ABS symbols?? I remember seeing something like this in the forum.
Nice thought not sure though! Vittal, if u did square it how do you see the solution panning out?

Wouldnt u still get 2 solutions? Let me know what your thoughts are.
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Re: Mod problem - anything wrong with this approach

by srisl11 » Thu Nov 27, 2008 4:08 pm
pbanavara wrote:is |x| < 1 ?

|x+1| = 2|x-1|
|x-3| != 0

Taking option 1:

x+1 = 2(x-1) or x+1 = 2(1-x) or -x-1 = 2(x-1) or -x-1 = 2(1-x)

Solving x=1/3 or x=3 Not sufficient

Option be x-3 !=0 or 3-x!=0 => x != 3 not sufficeint

Combining both x=3 which answers the question .. but the OA is different - what could be wrong here ..

Thanks,
Pradeep
IMO C

from 1 : x = 1/3 or 3 satisfy |x+1| = 2|x-1|
=> no other value of x (except for 1/3 or 3) satisfy the above equation.

from 2 : x != 3
=> x can be any value except 3

If we combine 1 and 2 , only x= 1/3 will satisfy both the equations

So IMO C

Or

we can use set theory too
Solution set for st 1 (values satisfying eq 1)= {1/3, 3}
Solution set for st 2(values satisfying eq 2) = {...-1,...-1/3,...0,...1/3....1,...2,.....2.9, 3.1....}
=> all numbers except 3

If we take the intersection of both sets we get {1/3} which is the solution set for the value which will satisfy both the equations
So x = 1/3
Please correct me if I'm wrong
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