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```Date: 03/18/2003 at 22:26:06
From: Nicole

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

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

(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.

(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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