I. u = √x, v = √(x - 12) u + v = 6 u² - v² = 12 (u - v)(u + v) = 12 6(u - v) = 12 u - v = 2 u = (6 + 2)/2 = 4 x = u² = 16 II. The left side of the equation is an increasing function, the right one is constant, so if the equation has a root, then it is the only one. By checking squares: 1, 4, 9, 16, we find 16 to be a solution.
rt(x-12)=6-rtx
x-12=36-12rtx+x
12rtx=48
rtx=4
x=16
I. u = √x, v = √(x - 12)
u + v = 6
u² - v² = 12
(u - v)(u + v) = 12
6(u - v) = 12
u - v = 2
u = (6 + 2)/2 = 4
x = u² = 16
II. The left side of the equation is an increasing function, the right one is constant, so if the equation has a root, then it is the only one. By checking squares: 1, 4, 9, 16, we find 16 to be a solution.