Inequality Problem Set

[GMAT Fever]

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!

[Sanjeev]

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

[GMAT Fever]

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!

[sanjeev]

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.

[RonPurewal]

http://www.manhattangmat.com/forums/abs ... t3648.html