Question

which solution to the equation \\(\frac{3}{a + 2} + \frac{2}{a} = \frac{4a - 4}{a^2 - 4}\\) is extraneous? \\(a = -2\\) \\(a = 4\\) neither \\(a = -2\\) nor \\(a = 4\\) \\(a = -2\\) and \\(a = 4\\)

Explanation:

Step1: Find Domain Restrictions

The original equation is \(\frac{3}{a + 2}+\frac{2}{a}=\frac{4a - 4}{a^{2}-4}\). First, factor the denominator \(a^{2}-4=(a + 2)(a - 2)\). The domain restrictions occur where denominators are zero: \(a+2
eq0\Rightarrow a
eq - 2\), \(a
eq0\), and \(a - 2
eq0\Rightarrow a
eq2\). So any solution equal to \(-2\), \(0\), or \(2\) is extraneous or not in the domain.

Step2: Solve the Equation

Multiply each term by the least common denominator \(a(a + 2)(a - 2)\) to eliminate the fractions:

$$
LATEXBLOCK0
$$

Factor the quadratic equation: \(a^{2}-2a - 8=(a - 4)(a + 2)=0\). So the solutions are \(a = 4\) and \(a=-2\).

Step3: Check for Extraneous Solutions

From the domain restrictions, \(a=-2\) makes the original denominators \(a + 2 = 0\) and \(a^{2}-4=0\), so \(a=-2\) is extraneous. \(a = 4\) does not make any denominator zero, so it is a valid solution.

Answer:

A. \(a=-2\)