Question 132124
Yes, it is called a rational equation because the word rational means fraction.  It is ALSO called, you guessed it, fractional equations.

I think this is your question:

(x^2)/(x^2 - 4) = x/(x + 2) - [2x/(2 - x)]

The first thing I would do is multiply the quantity (2 - x) times -1 to change 2 - x into x + 2.  Is this clear?

We now have this fraction:

(x^2)/(x^2 - 4) = x/(x + 2) - [2x/(x + 2)]

Do you see x^2 - 4?  That is your LCD.  You will multiply the LCD by every fraction on BOTH sides of the equation.

Why do that?

By doing this, you will REMOVE the fractions, which makes the problem a lot easier to play with.

Remember that x^ - 4 is called the difference of two perfect squares.

Recall from pre-algebra?

In other words, x^2 - 4 = (x - 2)(x + 2), recall?

(x^2)/x^2 - 4) times (x - 2)(x + 2) = x^2

x/x + 2) times (x - 2)(x + 2) = x(x - 2) = x^2 - 2x

-2x/(x + 2) times (x - 2)(x + 2) = -2x(x - 2) = -2x^2 + 4x

We now have a quadratic (NOT RATIONAL) equation and it is:

x^2 = x^2 - 2x - 2x^2 + 4x

Combine like terms on the right side of the equation to get:

x^2 = -x^2 + 2x

Bring the right side to the left side and equate to zero like this:

x^2 + x^2 - 2x = 0

2x^2 - 2x = 0

Factor the left side like this:

2x(x - 1) = 0

Equate EACH FACTOR to zero and solve for x.

2x = 0

x = 0/2

x = 0

NEXT:

x - 1 = 0

x = 1

We have two answers for x: 0 or 1.

How do we know which is correct?

REPLACE x with 0 and then replace x with 1 in the original rational equation.

If you get the same answer on BOTH sides of the equation, then you will know the true value(s) of x.

I will plug ZERO and simplify BUT you must PLUG 1 and simplify.

How about that?

0^2/0^2 - 4 = 0/0 + 2 - 2(0)/2 - 0

0/-4 = 0/2 - 0/2

0 = 0 - 0

0 = 0...It matches!!  Very nice!  So, this means that x = 0 is correct.

What about the other value for x?  How about x = 1?  Is that true?

I will let you plug that into the original rational equation and simplify.

Is this clear?