Question

which solution to the equation \\(\frac{3}{2g + 8} = \frac{g + 2}{g^2 - 16}\\) is extraneous? \\(g = -4\\) and \\(g = 16\\); \\(g = -4\\); \\(g = 16\\); neither \\(g = -4\\) nor \\(g = 16\\)

Explanation:

Step1: Factor Denominators

Factor \(2g + 8 = 2(g + 4)\) and \(g^2 - 16=(g + 4)(g - 4)\). The equation becomes \(\frac{3}{2(g + 4)}=\frac{g + 2}{(g + 4)(g - 4)}\).

Step2: Find Restrictions

Denominators can't be zero, so \(g
eq - 4\) and \(g
eq4\).

Step3: Solve the Equation

Multiply both sides by \(2(g + 4)(g - 4)\) (LCM of denominators):
\(3(g - 4)=2(g + 2)\)
Expand: \(3g-12 = 2g + 4\)
Subtract \(2g\) and add 12: \(g=16\).

Step4: Check Extraneous Solution

\(g = - 4\) is not in the solution (\(g = 16\)), but check \(g=-4\) in original equation: denominators \(2(-4)+8 = 0\) and \((-4)^2-16 = 0\), so \(g=-4\) makes denominators zero (undefined), so it's extraneous. \(g = 16\) is valid (denominators non - zero).

Answer:

B. \(g=-4\)