Register    Login    Search    Rss Feeds

Forums Home » Ask An Instructor » GMAT Math » Manhattan GMAT CAT Math


 Page 1 of 1 [ 5 posts ] 



  Print view Print view Previous topic Previous topic | Next topic Next topic
 

Inequality Problem Set

Author Message
GMAT Fever
 Post subject: Inequality Problem Set
 Post Posted: Tue Jul 01, 2008 10:40 pm 
Is |x| < 1 ?

(1) |x + 1| = 2|x – 1|

(2) |x – 3| > 0

I was able to reprhase the question to Is -1 < x < 1?

Then started with 2) because it seemed easier and got it to:
x – 3| > 0 x – 3 > 0 x > 3
|x – 3| > 0 3 – x > 0 x < 3

Which is clearly NS.

Then for 1).
Here is where I get confused.

I solved for two cases X<0
x + 1| = 2|x –1| -(x + 1) = 2(1 – x) x = 3

Then X>0
|x + 1| = 2|x –1| x + 1 = 2(x – 1) x = 3

So from here I got 3 for both answers and thought that this S so I picked A.

Apparently this is incorrect. Can someone explain where I went wrong and what I am missing? Thanks!
Top 
Sanjeev
 Post subject:
 Post Posted: Wed Jul 02, 2008 8:08 pm 
Hi,

When you have two equal mod values,the only thing you know is that the scalar value of LHS = scalar value of RHS.

In the below case:-
(1) |x + 1| = 2 |x -1 |


I dont think you can solve it this way:-
X<0
x + 1| = 2|x –1| -(x + 1) = 2(1 – x) x = 3

Then X>0
|x + 1| = 2|x –1| x + 1 = 2(x – 1) x = 3


One method which i follow is take the Square of both LHS and RHS , since we know that If there are two equal numbers irrespective of their sign then their square would also be equal.

So,

(|x +1|)^2 = 4 (|x -1| )^2
=> x^2 +2x`+1 = 4x^2 -8x + 4
=> Solving this u get 3x^2 -10x +3 = 0
=> u get two values x = 1/3 and x = 3


(2) |x - 3| > 0
This gives x cannot be equal to 3


Combine (1) and (2) , u get x = 1/3


So i think answer is (c) ..

Is that right?


Thanks
Top 
GMAT Fever
 Post subject:
 Post Posted: Wed Jul 02, 2008 10:08 pm 
Sanjeev wrote:

One method which i follow is take the Square of both LHS and RHS , since we know that If there are two equal numbers irrespective of their sign then their square would also be equal.

So,

(|x +1|)^2 = 4 (|x -1| )^2
=> x^2 +2x`+1 = 4x^2 -8x + 4
=> Solving this u get 3x^2 -10x +3 = 0
=> u get two values x = 1/3 and x = 3




Sanjeev - You are correct! The answer is C, good job!

As for your method above, I really like it! Just a few questions:

-First a remedial question how did you factor 3x^2 -10x +3 = 0 to get x = 1/3 and x = 3? When I attempt to factor the equation I get (x^2 - 10/3x + 1) - and the factors you suggested dont work...am I missing something? Did you factor a diff way?
-Should the squaring method only be applied if the numbers are equal?
-Is there a case when two abs value expressions are set equal and this method doesnt work? If so can you provide example?

Thanks for your help!
Top 
sanjeev
 Post subject:
 Post Posted: Mon Jul 07, 2008 1:06 am 
Hi,

Thanks.

Squaring will only work when LHS = RHS. For instance lets take a case
-3 < -2 is a fact
however when you square , we get 9 < 4, which is not correct.

I follow a method called "Middle Term Break" to factor.Its easy , but it would require practice to master it.
Take an example of quadratic equation
ax^2 + bx + c = 0
=Split b in two terms say b1 and b2 such that b1 + b2 = b and b1 * b2 = c * a

In our case , it is 3x^2 - 10x +3 = 0
Here a= 3 , b = -10 and c = 3 ,
So here b1 and b2 would be -9 and -1. To verify check b1 + b2 = -10 and b1 * b2 = 9

So , above equation can be written as 3x^2 -9x -x +3 = 0
=>3x(x - 3) -1(x - 3) = 0
=> (3x- 1) (x-3) = 0
=> x = 1/3,3

I dont think there are any exception for squaring when the LHS = RHS.
Top 
RonPurewal
 Post subject:
 Post Posted: Fri Jul 18, 2008 7:08 pm 
Offline
ManhattanGMAT Staff


Posts: 7146
http://www.manhattangmat.com/forums/abs ... t3648.html
Top 
Display posts from previous:  Sort by  
 
 Page 1 of 1 [ 5 posts ] 





Who is online

Users browsing this forum: No registered users and 0 guests

 
 

 
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to: