\( \sqrt{16 x+41}=x+6 \) The solution set is

To solve the equation \( \sqrt{16x + 41} = x + 6 \), start by squaring both sides to eliminate the square root: \[ 16x + 41 = (x + 6)^2 \] Expanding the right side: \[ 16x + 41 = x^2 + 12x + 36 \] Rearranging gives: \[ 0 = x^2 - 4x - 5 \] Now, factor the quadratic: \[ 0 = (x - 5)(x + 1) \] This yields the solutions \( x = 5 \) and \( x = -1 \). However, we must check these against the original equation: 1. For \( x = 5 \): \[ \sqrt{16(5) + 41} = \sqrt{80 + 41} = \sqrt{121} = 11 \] \[ 5 + 6 = 11 \quad \text{(valid)} \] 2. For \( x = -1 \): \[ \sqrt{16(-1) + 41} = \sqrt{-16 + 41} = \sqrt{25} = 5 \] \[ -1 + 6 = 5 \quad \text{(valid)} \] Thus, the solution set is \( \{5, -1\} \).