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Quadratic Inequalities

Date: 03/18/2003 at 22:26:06
From: Nicole
Subject: Quadratic inequalities

When I solve (x^2 + 1)/4 >= (x + 2)/2 , of my final two answers, only 
one works. The other should have the sign turned.

I multiplied both sides by 4 and came up with x^2 + 1 > 2x + 4
I brought everything to the left and got x^2 - 2x - 3 > 0
After factoring I got (x - 3)(x + 1) > 0
So I did x - 3 > 0 and x + 1 > 0
Solving those two I get x > 3 and x > -1.  

However:
Checking a number in the original problem that is greater than 3 
works BUT checking a number greater than -1 (like 0) doesn't work; 
putting 0 in the original gives me 1/4 > 2/2 which is not.  

What am I missing here?


Date: 03/20/2003 at 22:58:41
From: Doctor Dotty
Subject: Re: Quadratic inequalities

Hi Nicole,

Thanks for the question!

You are completely correct that the original inequality is equivalent 
to 

  (x - 3)(x + 1) >= 0

(where '>=' means 'greater than or equal to').

I'll address what you 'missed' first, and then show you a couple of
easier methods to solve this problem. 

What you 'missed is that 

  (x - 3)(x + 1) >= 0 

is not equivalent to 

  (x - 3) >= 0  AND  (x + 1) >= 0

My guess is that you used the same reasoning that you'd use for a
quadratic equation, for example, 

    (x - 3)(x + 1) = 0

The reason it would be a valid method in this case is because if you 
divide both sides by (x + 1), you get:

           (x - 3) = 0

If you try to do this with an inequality, you run into problems. If 
you divide both sides by (x + 1) when (x + 1) is negative, then the 
sign switches direction. So to solve it in this way, you need two 
cases, one where (x + 1) >= 0 and one where it is not. This is 
fairly difficult and is certainly not the most efficient method of 
solving.

The easiest way to solve this sort of question is to forget all the 
above and picture what is actually happening.

Let's use your example:

 (x - 3)(x + 1) >= 0

To decide what it looks like, we firstly need to find the points 
where it crosses the x-axis (known as roots). To do that we _equal_ 
it to 0.

    (x - 3)(x + 1) = 0

So it crosses the x-axis at x = 3 or -1, however we don't know in 
which direction they go through.

There are two ways to find out, drawing a graph or checking numbers.

As the x^2 is positive, the graph looks a bit like:

         *       |          *
          *      |         *
           *     |        *
            *    |       *
             *   |      *
     ---------*--|-----*-------
            -1  *|    * 3
                 |* *
                 |
                 |
                 |

We can see that the curve is >= 0 when x is <= -1 or >= 3.

Alternatively, you could use the method you used at the end and try 
some values on either side or each root:

  when x is -2, (x - 3)(x + 1) is positive

  when x is -1, we know that (x - 3)(x + 1) is 0

  when x is 0, (x - 3)(x + 1) is negative

  when x is 3, we know that (x - 3)(x + 1) is 0
 
  when x is 4, (x - 3)(x + 1) is positive

We can see that (x - 3)(x + 1) is >= 0 when x is <= -1 or >= 3.

Does that make sense? 

If I can help any more with this problem or any other, please write 
back.

- Doctor Dotty, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra
High School Equations, Graphs, Translations
High School Polynomials

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