Calculus optimization help

Suppose that light reflects off a mirror to get from point A to point B as indicated in the figure. Assuming a constant velocity of light, we can minimize time by minimizing the distance traveled. Find the point on the mirror that minimizes the distance traveled. Show that the angles in the figures are equal (the angle of incidence equals the angle of reflection).

http://highered.mcgraw-hill.com/sites/dl… (#28) Please help, I just can't quite figure out where to start with this..

Since the speed of light is invariant, all we really have to do is find the minimum distance and we'll find the minimum time it takes for light to travel from A to B

Notice that we have 2 right triangles. 1 triangle has a side of 2 and another side of x. The other triangle has a side of 1 and another side of (4 - x). The path that the light travels is therefore the sum of the hypotenuses of the 2 triangles:

D = sqrt(2^2 + x^2) + sqrt(1^2 + (4 - x)^2)

Find when dD/dx = 0

D = (4 + x^2)^(1/2) + (1 + (4 - x)^2)^(1/2)
D = (4 + x^2)^(1/2) + (1 + 16 - 8x + x^2)^(1/2)
D = (4 + x^2)^(1/2) + (17 - 8x + x^2)^(1/2)

dD/dx = (1/2) * (2x) * (4 + x^2)^(-1/2) + (1/2) * (-8 + 2x) * (17 - 8x + x^2)^(-1/2)
dD/dx = x / sqrt(4 + x^2) + (x - 4) / sqrt(x^2 - 8x + 17)

dD/dx = 0
0 = x / sqrt(4 + x^2) + (x - 4) / sqrt(x^2 - 8x + 17)
(4 - x) / sqrt(x^2 - 8x + 17) = x / sqrt(4 + x^2)
(4 - x) * sqrt(4 + x^2) = x * sqrt(x^2 - 8x + 17)

Square both sides

(4 - x)^2 * (4 + x^2) = x^2 * (x^2 - 8x + 17)
(16 - 8x + x^2) * (4 + x^2) = x^4 - 8x^3 + 17x^2
64 + 16x^2 - 32x - 8x^3 + 4x^2 + x^4 = x^4 - 8x^3 + 17x^2
x^4 - 8x^3 + 20x^2 - 32x + 64 = x^4 - 8x^3 + 17x^2
20x^2 - 32x + 64 = 17x^2
3x^2 - 32x + 64 = 0
x = (32 +/- sqrt(32^2 - 4 * 3 * 64)) / 6
x = (32 +/- sqrt(32^2 - 4 * 3 * 2 * 32)) / 6
x = (32 +/- sqrt(32 * (32 - 4 * 3 * 2))) / 6
x = (32 +/- sqrt(32 * (32 - 24)) / 6