Absolute Value: Two Methods

Date: 07/03/98 at 10:46:37
From: Presley
Subject: Absolute value

Hello Dr. Math,

I have a problem I'm curious about:

Solve for x: |x+3| < |x-2|

I wanted to square both sides, but I am scared that it is not correct. 
Can you show me how to solve this equation?

Date: 07/04/98 at 05:17:12
From: Doctor Anthony
Subject: Re: Absolute value

For this particular problem you could square both sides, but that 
technique is not always applicable, for example 

|x-2| + |x+5| - |x-3| > |x+13|

 So in the example you gave

    |x+3| < |x-2|   squaring both sides

    x^2 + 6x + 9 < x^2 - 4x + 4

           10x < -5

             x < -1/2

Alternative method.
-------------------

Critical values are -3 and +2, so we consider 3 separate situations:

x < -3   Inequality can be written  -(x+3) < -(x-2)

                                    -x - 3 < -x + 2
                                       
                                       - 3 < 2   (always true)

So the inequality is satisfied for all x < -3

Second region is  -3 < x < 2   Inequality can be written

                                x+3 < -(x-2)

                                x + 3 < - x + 2

                                   2x < -1

                                    x < -1/2

So the inequality is true for all x < -1/2

Third region is  x > 2.  Inequality can be written

                             x+3 < x-2

                               3 < -2   (never true)

So no values of x satisfy the inequality in the region x > 2

Combining all three regions we see that the inequality is satisfied 
when:
           x < -1/2

This is the same result as before, and in this example the first 
method is clearly to be preferred.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/