Calculate the real number x>=4, when 2, x-4, x are the consecutive terms of a geometric progression.

If 2, x-4, x are consecutive terms of the geometric progression, that means that:

x-4 = 2q, where q is the ratio

We'll divide by 2 and we'll get:

q = (x-4)/2 (1)

x = q*(x-4)

We'll divide by (x-4)

q = x/(x-4) (2)

From (1) and (2) it results:

(x-4)/2 = x/(x-4)

We'll cross multiplying:

(x-4)^2 = 2x

We'll expand the square:

x^2 - 8x + 16 = 2x

We'll move all terms to one side:

x^2 - 8x + 16 - 2x = 0

x^2 - 10x + 16 = 0

We'll apply the quadratic formula:

x1 = [10+sqrt(100-64)]/2

x1 = (10+6)/2

x1 = 16/2

x1 = 8

x2 = (10-6)/2

x2 = 2

Because, from the enunciation, x has to be more than 4, or at least 4, the second solution is not convenient. So, the only solution of the equation is x = 8.