To solve x = 1+(1+x^(1/2))^(1/2).

Subtract 1 from both sides:

x-1 = (1+x^(1/2))^(1/2)

We square both sides:

(x-1)^2 = (1+x^1/2).

(x-1)^2 -1 = x^(1/2).

We square both sides:

{(x-1)^2 -1 }^2 = x.

{x^2 -2x}^2= x.

x^4-4x^3 +4x^2 -x = 0

x^3 - 4x^2 +4x-1 = 0

x^3-1 -4x(x-1) = 0

(x-1){ x^2+x+1)-4x(x-1) = 0.

(x-1) (x^2-3x+1) = 0

x-1 = 0, or (x^2-3x+1) = 0.

x= 1, or x = {3+sqrt(9-4)}/2 , Or x = {1+sqrt(5)}/2

x = 1, x = (3+sqr5)/2 , x= (3-sqrt5)/2.

First, we'll impose the constraint of existence of square root;

x>=0

1 + sqrtx>=0

We'll subtract 1 both sides:

x - 1 = sqrt[1 + sqrt(x)]

We'll raise to square both sides, to eliminate the square root from the right side:

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

We'll expand the square:

x^2 - 2x + 1 = 1 + sqrtx

We'll eliminate like terms:

x^2 - 2x = sqrt x

We'll raise to square:

(x^2 - 2x)^2 = x

[x(x - 2)]^2 - x = 0

We'll factorize by x:

x[x(x-2)^2 - 1] = 0

x1 = 0

x(x-2)^2 - 1 = 0

We'll expand the square:

x^3 - 4x^2 + 4x - 1 = 0

We'll factorize the middle terms by -4x:

(x^3 - 1) - 4x(x - 1) = 0

We'll re-write the difference of cubes;

x^3 - 1 = (x - 1)(x^2 + x + 1)

(x - 1)(x^2 + x + 1) - 4x(x - 1) = 0

We'll factorize by (x-1):

(x-1)(x^2 + x + 1 - 4x) = 0

x - 1 = 0

x2 = 1

x^2 - 3x + 1 = 0

We'll apply the quadratic formula:

x3 = [3+sqrt(9 - 4)]/2

x3 = (3 + sqrt5)/2

x4 = (3 - sqrt5)/2

**The solutions of the equation are: {0 ; (3 - sqrt5)/2 ; 1 ; (3 + sqrt5)/2}.**