To determine which equation has exactly one solution, let's analyze each option step by step:
A) \( 8 + 3x^2 = 13 + x \)
Rearranging gives: \[ 3x^2 - x - 5 = 0 \] This is a quadratic equation. To determine the number of solutions, we can use the discriminant \( D = b^2 - 4ac \): \[ a = 3, b = -1, c = -5 \] \[ D = (-1)^2 - 4 \cdot 3 \cdot (-5) = 1 + 60 = 61 \] Since \( D > 0 \), this equation has two solutions.
B) \( 3 + 12x^2 = 6x + 15 \)
Rearranging gives: \[ 12x^2 - 6x - 12 = 0 \] Dividing through by 6: \[ 2x^2 - x - 2 = 0 \] Let's find the discriminant: \[ a = 2, b = -1, c = -2 \] \[ D = (-1)^2 - 4 \cdot 2 \cdot (-2) = 1 + 16 = 17 \] Since \( D > 0 \), this equation also has two solutions.
C) \( 10 + 15z^5 = 3x + 10 \)
Rearranging gives: \[ 15z^5 - 3x = 0 \] To have a unique solution, let's set \( z = 0 \): \[ 15(0)^5 - 3x = 0 \] This means: \[ -3x = 0 \implies x = 0 \] So it depends on if \( z \) could equal more values. Since \( z^5 \) could take many values, this does not lead to one solution.
D) \( 4 - 20 = x - 5 \)
Simplifying gives: \[ -16 = x - 5 \] So, \[ x = -11 \] This is indeed a unique solution.
After analyzing all equations, the equation that has exactly one solution is:
D) \( 4 - 20 = x - 5 \)