Solve the equation x=1+square root(1+square root x).

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}.