Solve the equation for x square root(1-square root(x^4-x))=x-1

First, we'll raise to square both sides, to eliminate the
square root from the left side:

1-square root(x^4-x)=
(x-1)^2

We'll eliminate like
terms:

1-square root(x^4-x)= x^2 - 2x + 1 -square
root(x^4-x)= x^2 - 2x

We'll raise to square both sides
again: x^4 - x = x^4 - 4x^3 + 4x^2

We'll eliminate x^4 both
sides: 4x^3 - 4x^2 - x = 0

We'll factorize by x: x(4x^2 -
4x - 1) = 0

We'll put x = 0. 4x^2 - 4x - 1 = 0 We'll apply
quadratic formula:

x1 = [4+sqrt(16+16)]/16 x1 = 4(1 +
sqrt2)/16 x1 = (1 + sqrt2)/4 x2 = (1 - sqrt2)/4

Since the
range of admissible values for x is: {0}U[(1 + sqrt2)/4 ; +infinite), we'll reject the
second value for x.

The solutions of the
equation are: {0 ; (1 + sqrt2)/4 }.