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 away from the point \( (5, -8) \), follow these steps: 1. **Identify the Coordinates on the \( x \)-axis:** Points on the \( x \)-axis have the form \( (x, 0) \). 2. **Use the Distance Formula:** The distance \( d \) between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \] Plugging in the given points \( (5, -8) \) and \( (x, 0) \): \[ 16 = \sqrt{(x - 5)^2 + (0 - (-8))^2} \] 3. **Simplify the Equation:** \[ 16 = \sqrt{(x - 5)^2 + 64} \] Square both sides: \[ 256 = (x - 5)^2 + 64 \] Subtract 64 from both sides: \[ (x - 5)^2 = 192 \] 4. **Solve for \( x \):** \[ x - 5 = \pm \sqrt{192} = \pm 8\sqrt{3} \] Therefore: \[ x = 5 \pm 8\sqrt{3} \] 5. **Write the Points:** The corresponding points on the \( x \)-axis are: \[ (5 - 8\sqrt{3}, 0) \quad \text{and} \quad (5 + 8\sqrt{3}, 0) \] **Final Answer:** \((5-8\sqrt{3}, 0),\ (5+8\sqrt{3}, 0)\)