Answered

Which solution to the equation $\frac{3}{2 g+8}=\frac{g+2}{g^2-16}$ is extraneous?


A$g=-4$
B$g=-4$ and $g=16$
Cneither $g=-4$ nor $g=16$
D$g=16$

Verified Answer :

  • Analyze the denominators: Factor \(2g+8\) as \(2(g+4)\) and \(g^2-16\) as \((g-4)(g+4)\).
  • Determine restricted values: The equation is undefined where denominators equal zero, specifically \(g = -4\) and \(g = 4\).
  • Test the candidates: Substituting \(g = -4\) into the equation makes the denominators zero, identifying it as an extraneous solution.
  • Verify \(g=16\): Substituting \(g = 16\) results in both sides of the equation being equal, so it is a valid solution.
  • Final Answer: The extraneous solution is \(\boxed{g = -4}\).

Explanation

  1. Identifying Restricted Values

To identify the extraneous solutions of the rational equation \(\frac{3}{2g+8} = \frac{g+2}{g^2-16}\), we must first determine the values of \(g\) for which the equation is undefined. An extraneous solution arises when a value makes the denominator of any rational expression in the equation equal to zero.

The denominators are:

  1. \(2g + 8\)
  2. \(g^2 - 16\)

Factoring these expressions:

  1. \(2g + 8 = 2(g + 4)\)
  2. \(g^2 - 16 = (g - 4)(g + 4)\)

The equation is undefined when any of these factors is zero:

  • \(2(g+4) = 0 \implies g = -4\)
  • \(g - 4 = 0 \implies g = 4\)
  • \(g + 4 = 0 \implies g = -4\)

Thus, the values \(g = -4\) and \(g = 4\) make the equation undefined. Any potential solution equal to either of these values must be extraneous.

  1. Evaluating Potential Solutions

We are given potential solutions \(g = -4\) and \(g = 16\).

  • For \(g = -4\): Substituting this into the denominators gives \(2(-4) + 8 = 0\) and \((-4)^2 - 16 = 0\). Since the denominators are zero, the expression is undefined at \(g = -4\). Thus, \(g = -4\) is an extraneous solution.
  • For \(g = 16\): Substituting this into the equation:

Left-hand side: \(\frac{3}{2(16) + 8} = \frac{3}{32 + 8} = \frac{3}{40}\)
Right-hand side: \(\frac{16 + 2}{16^2 - 16} = \frac{18}{256 - 16} = \frac{18}{240} = \frac{3}{40}\)
Since both sides are equal (\(\frac{3}{40} = \frac{3}{40}\)), \(g = 16\) is a valid solution, not an extraneous one.

  1. Conclusion

Based on our analysis, the only extraneous solution provided in the list is \(g = -4\). The value \(g = 16\) is a valid solution to the given equation. Therefore, the correct option is \(g = -4\).

Examples

Extraneous solutions occur in many real-world applications where rational functions model relationships. For instance, if you are designing a system where the time taken to complete a task depends on the speed or flow rate (\(t = \frac{d}{v}\)), you must ensure that your model never sets the speed or denominator to zero, which would represent a physical impossibility like a stationary object or a blocked pipe. Identifying extraneous solutions helps you filter out mathematically "valid-looking" results that are physically meaningless, ensuring the reliability of your model.

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