Inequalities and Absolute Value - Case Method

Date: 09/01/98 at 05:46:18
From: Lex
Subject: Inequalities and Absolute Value

What is the case method? How does it apply to inequalities with 
absolute values?

Please solve:

   | x + 4 | <= 2
   | x - 3 | <= | x |

Date: 09/01/98 at 07:11:52
From: Doctor Anthony
Subject: Re: Inequalities and Absolute Value

For |x+4| <= 2, there are two ways you could solve this.  

Method(1): Square both sides:

   x^2 + 8x + 16 <= 4
   x^2 + 8x + 12 <=0
      (x+2)(x+6) <=0

You can think of a parabola which crosses the x axis at -6 and -2 and 
is below the x axis between these values. So y is negative between 
x = -6 and -2. Thus, the inequality is satisfied for -6 <= x <= -2.

Method (2):  |x+4| <= 2

The critical value is when x = -4 and you must consider separately 
cases when x <= -4 and when x >= -4. This is the case method.

When x <= -4 we could write the equation:

   -(x+4) <= 2
    -x-4  <= 2
      -6  <= x

So in this case x >= -6 but less than -4.

When x >= -4 we could write the equation:

   x+4 <= 2
     x <= -2

So in this case x lies between -4 and -2.

Combining these two situations we have, as with method (1):

   -6 <= x <= -2

Although method (1) looks better than method (2) for this problem, 
there are situations where method (1) can't be used. For example, in 
problems like that shown below, method (1) would be very difficult to 
apply.

  |x-4| + |3x+2| + |x+7| >= 6
 
For | x - 3 | <= | x |, using method(1) we require:

   x^2 - 6x + 9 <= x^2
       - 6x + 9 <= 0
              9 <= 6x
            3/2 <= x

So here we require  x >= 3/2.

As an exercise you should try this problem using method (2). The 
critical values are x = 3 for |x-3| and x = 0 for |x|.

- Doctor Anthony, The Math Forum
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