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Determining the Values For Which a Rational Expression is Undefined

Determining the Values For Which a Rational Expression is Undefined

When we work with a numerical fraction, it is easy to avoid dividing by zero, because we can see the number in the denominator. In order to avoid dividing by zero in a rational expression, we must not allow values of the variable that will make the denominator be zero.

If the denominator is zero, the rational expression is undefined. The numerator of a rational expression may be 0—but not the denominator.

So before we begin any operation with a rational expression, we examine it first to find the values that would make the denominator zero. That way, when we solve a rational equation for example, we will know whether the algebraic solutions we find are allowed or not.

Determine the Values for Which a Rational Expression is Undefined.

  1. Set the denominator equal to zero.
  2. Solve the equation in the set of reals, if possible.

Example

Determine the values for which the rational expression is undefined:

  1. \(\frac{9y}{x}\)
  2. \(\frac{4b-3}{2b+5}\)
  3. \(\frac{x+4}{{x}^{2}+5x+6}\)

Solution

The expression will be undefined when the denominator is zero.

  1.  

    \(\begin{array}{cccc}& & & \hfill \phantom{\rule{8em}{0ex}}\frac{9y}{x}\hfill \\ \begin{array}{c}\text{Set the denominator equal to zero. Solve}\hfill \\ \text{for the variable.}\hfill \end{array}\hfill & & & \hfill \phantom{\rule{8em}{0ex}}x=0\hfill \\ & & & \hfill \phantom{\rule{8em}{0ex}}\frac{9y}{x}\phantom{\rule{0.2em}{0ex}}\text{is undefined for}\phantom{\rule{0.2em}{0ex}}x=0.\hfill \end{array}\)

  2.  

    \(\begin{array}{cccc}& & & \hfill \phantom{\rule{6em}{0ex}}\frac{4b-3}{2b+5}\hfill \\ \begin{array}{c}\text{Set the denominator equal to zero. Solve}\hfill \\ \text{for the variable.}\hfill \end{array}\hfill & & & \hfill \phantom{\rule{6em}{0ex}}\begin{array}{ccc}\hfill 2b+5& =\hfill & 0\hfill \\ \hfill 2b& =\hfill & -5\hfill \\ \hfill b& =\hfill & -\frac{5}{2}\hfill \end{array}\hfill \\ & & & \hfill \phantom{\rule{6em}{0ex}}\frac{4b-3}{2b+5}\phantom{\rule{0.2em}{0ex}}\text{is undefined for}\phantom{\rule{0.2em}{0ex}}b=-\frac{5}{2}.\hfill \end{array}\)

  3.  

    \(\begin{array}{cccc}& & & \hfill \phantom{\rule{4em}{0ex}}\frac{x+4}{{x}^{2}+5x+6}\hfill \\ \begin{array}{c}\text{Set the denominator equal to zero. Solve}\hfill \\ \text{for the variable.}\hfill \end{array}\hfill & & & \hfill \phantom{\rule{4em}{0ex}}\begin{array}{ccc}\hfill {x}^{2}+5x+6& =\hfill & 0\hfill \\ \hfill \left(x+2\right)\left(x+3\right)& =\hfill & 0\hfill \\ \hfill x+2=0\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}x+3& =\hfill & 0\hfill \\ \hfill x=-2\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}x& =\hfill & -3\hfill \end{array}\hfill \\ & & & \hfill \phantom{\rule{4em}{0ex}}\frac{x+4}{{x}^{2}+5x+6}\phantom{\rule{0.2em}{0ex}}\text{is undefined for}\phantom{\rule{0.2em}{0ex}}x=-2\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}x=-3.\hfill \end{array}\)

Saying that the rational expression \(\frac{x+4}{{x}^{2}+5x+6}\) is undefined for \(x=-2\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}x=-3\) is similar to writing the phrase “void where prohibited” in contest rules.

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