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Question about rational expressions and their domains

  1. May 13, 2013 #1
    Hi there. I am currently taking "College Math 1" at the local CC and I have encountered something that confuses me regarding rational expressions and their domains. The definition given by the textbook for rational expressions is:

    "the set of real numbers for which an algebraic expression is defined is the domain of the expression. Because rational expressions indicate division and division by zero is undefined, we must exclude numbers from a rational expressions domain that make the denominator zero."

    What I gather from that is that I am looking for any numbers that make the denominator zero? Say I have a simple rational expression such as x2 + 6x + 5/x2-25 and I am trying to simplify it. After factoring I get x2 + 6x/x2-25 --> (x+1)(x+5)/(x+5)(x-5) --> x+1/x-5, x != -5, x != 5. Using the definition from the textbook for rational expressions, I see how x=5 would be 5-5=0 which makes the expression undefined. My question is basically where does x != -5 come from? I only see why x != 5 because then the denominator would be zero.

    Any answers are appreciated.
  2. jcsd
  3. May 13, 2013 #2
    You seem to think that

    [tex]\frac{(x+1)(x+5)}{(x-5)(x+5)} = \frac{x+1}{x-5}[/tex]

    This is a common mistake. But it is not true. Why not? If we set ##-5## in the left-hand side then it is not defined (because of division by 0). If we set ##-5## in the right-hand side, then we got something that is perfectly well-defined.

    What is true is that the above equality holds for all values ##x## for which both sides are defined. So for all ##x## which are not ##\pm 5##, the equality holds. But in general, it does not hold.
  4. May 24, 2013 #3
    It's called a missing solution. You can read about them here: http://en.wikipedia.org/wiki/Extraneous_and_missing_solutions. Do you see that when you divide by x+5 there is the possibility that you are dividing by zero when x = -5? Since x^2-25 is a second degree equation you know there are two solutions or possible zeros that make the original equation undefined(5 and -5). Since -5 is not a solution to the second equation they are technically not equal.
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