Which of the following is an extraneous solution of \( \sqrt{4 x+41}=x+5 ? \) \( \begin{array}{l}x=-8 \\ x=-2 \\ x=2 \\ x=8\end{array} \) \( \begin{array}{l}x=\end{array} \)

To determine the extraneous solution of the equation \( \sqrt{4x+41} = x+5 \), we can start by squaring both sides to eliminate the square root. This gives us \( 4x + 41 = (x + 5)^2 \). Expanding the right side, we get \( 4x + 41 = x^2 + 10x + 25 \). Rearranging everything to one side leads to \( x^2 + 6x - 16 = 0 \). Now, we can factor this quadratic equation: \( (x - 2)(x + 8) = 0 \). Therefore, the possible solutions are \( x = 2 \) and \( x = -8 \). We must check which, if any, are extraneous by substituting back into the original equation. 1. For \( x = 2 \): \( \sqrt{4(2) + 41} = \sqrt{8 + 41} = \sqrt{49} = 7 \) \( 2 + 5 = 7 \), so this solution is valid. 2. For \( x = -8 \): \( \sqrt{4(-8) + 41} = \sqrt{-32 + 41} = \sqrt{9} = 3 \) \( -8 + 5 = -3 \), which is not equal, therefore this is an extraneous solution. So, the extraneous solution is \( x = -8 \).