# Equation .Find the solution of the square root equation x=1+square root(1+square root x) .

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We need to solve x=1+square root(1+square root x)

x = 1 + sqrt (1 + sqrt x)

=> x - 1 = sqrt (1 + sqrt x)

square the sides

=> x^2 + 1 - 2x = 1 + sqrt x

=> x^2 - 2x = sqrt x

square the sides

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

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

=> x(x^3 - 4x^2 + 4x - 1) = 0

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

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

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

The roots of x^2 - 3x + 1 = 0 are

3/2 + sqrt (9 - 4)/2 and 3/2 - sqrt (9 - 4)/2

**The roots of the equation are {0 , 1 , 3/2 + (sqrt 5)/2 , 3/2 - (sqrt 5)/2} **

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

x>=0

1 + sqrtx>=0

We'll subtract 1 both sides to isolate to the right side the square root:

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 write the difference of cubes as a product:

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