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

To determine the extraneous solution of the equation \( \frac{3}{2 g+8}=\frac{g+2}{g^{2}-16} \), we first simplify both sides. 1. **Factor the denominator on the right side**: The expression \( g^2 - 16 \) can be factored as \( (g - 4)(g + 4) \). Now our equation looks like: \[ \frac{3}{2g + 8} = \frac{g + 2}{(g - 4)(g + 4)} \] 2. **Cross-multiply**: \[ 3(g - 4)(g + 4) = (g + 2)(2g + 8) \] 3. **Expand both sides**: - The left side becomes: \[ 3(g^2 - 16) = 3g^2 - 48 \] - The right side is: \[ (g + 2)(2g + 8) = 2g^2 + 8g + 4g + 16 = 2g^2 + 12g + 16 \] 4. **Set the equation**: \[ 3g^2 - 48 = 2g^2 + 12g + 16 \] 5. **Move all terms to one side**: \[ 3g^2 - 2g^2 - 12g - 16 - 48 = 0 \] Which simplifies to: \[ g^2 - 12g - 64 = 0 \] 6. **Factor the quadratic**: We will look for two numbers that multiply to \(-64\) and add to \(-12\). These numbers are \(-16\) and \(4\). So we can factor as: \[ (g - 16)(g + 4) = 0 \] 7. **Find the solutions**: This gives us: \[ g - 16 = 0 \quad \Rightarrow \quad g = 16 \] \[ g + 4 = 0 \quad \Rightarrow \quad g = -4 \] 8. **Identify extraneous solutions**: Now we must check if either \( g = 16 \) or \( g = -4 \) causes any division by zero in the original equation. - For \( g = 16 \): \[ 2g + 8 = 2(16) + 8 = 32 + 8 = 40 \quad (\text{valid}) \] \[ g^2 - 16 = 16^2 - 16 = 256 - 16 = 240 \quad (\text{valid}) \] - For \( g = -4 \): \[ 2g + 8 = 2(-4) + 8 = -8 + 8 = 0 \quad (\text{undefined!}) \] \[ g^2 - 16 = (-4)^2 - 16 = 16 - 16 = 0 \quad (\text{undefined!}) \] Thus, the extraneous solution is \( g = -4 \) because it makes the left side undefined.