Find all points on the \( x \)-axis that are 16 units from the point \( (5,-8) \). All points on the \( x \)-axis that are 16 units from the point \( (5,-8) \) are (Simplify your answer. Type an ordered pair. Use a comma to separate answers as needed.)

To find all points on the \( x \)-axis that are 16 units from the point \( (5, -8) \), we can start by using the distance formula. Points on the \( x \)-axis have coordinates of the form \( (x, 0) \). The distance \( d \) between the points \( (x, 0) \) and \( (5, -8) \) is given by: \[ d = \sqrt{(x - 5)^2 + (0 + 8)^2} \] We set this equal to 16: \[ \sqrt{(x - 5)^2 + 8^2} = 16 \] Squaring both sides: \[ (x - 5)^2 + 64 = 256 \] Subtracting 64: \[ (x - 5)^2 = 192 \] Taking the square root: \[ x - 5 = \pm \sqrt{192} \] Since \( \sqrt{192} = 8\sqrt{3} \), we have: \[ x - 5 = 8\sqrt{3} \quad \text{or} \quad x - 5 = -8\sqrt{3} \] Solving for \( x \): 1. \( x = 5 + 8\sqrt{3} \) 2. \( x = 5 - 8\sqrt{3} \) Thus, all points on the \( x \)-axis that are 16 units from the point \( (5, -8) \) are: \[ (5 + 8\sqrt{3}, 0), (5 - 8\sqrt{3}, 0) \]