Find all points on the \( x \)-axis that are 16 units from the point \( (5,-8) \).

To find all points on the \( x \)-axis that are 16 units from the point \( (5, -8) \), we start with a generic point on the \( x \)-axis, which can be represented as \( (x, 0) \). The distance between the point \( (x, 0) \) and the point \( (5, -8) \) can be calculated using the distance formula: \[ d = \sqrt{(x - 5)^2 + (0 - (-8))^2} \] Simplifying this, we find: \[ d = \sqrt{(x - 5)^2 + 8^2} = \sqrt{(x - 5)^2 + 64} \] We want this distance to equal 16 units. Therefore, we set up the equation: \[ \sqrt{(x - 5)^2 + 64} = 16 \] Next, we square both sides to eliminate the square root: \[ (x - 5)^2 + 64 = 16^2 \] \[ (x - 5)^2 + 64 = 256 \] Now, we subtract 64 from both sides: \[ (x - 5)^2 = 256 - 64 \] \[ (x - 5)^2 = 192 \] Taking the square root of both sides gives us: \[ x - 5 = \pm \sqrt{192} \] We can simplify \( \sqrt{192} \): \[ \sqrt{192} = \sqrt{64 \times 3} = 8\sqrt{3} \] Thus, we have: \[ x - 5 = 8\sqrt{3} \quad \text{or} \quad x - 5 = -8\sqrt{3} \] Solving for \( x \): 1. For \( x - 5 = 8\sqrt{3} \): \[ x = 5 + 8\sqrt{3} \] 2. For \( x - 5 = -8\sqrt{3} \): \[ x = 5 - 8\sqrt{3} \] Therefore, the two points on the \( x \)-axis that are 16 units from the point \( (5, -8) \) are: \[ \boxed{(5 + 8\sqrt{3}, 0) \text{ and } (5 - 8\sqrt{3}, 0)} \]