square root(x+square root(1-x))+square rootx=1. What is x?

First, we'll impose the constraints of existence of square
roots:

x>=0

1 -
x>=0

x=<1

The
range of admissible values for x is [0 ; 1].

We'll solve
the equation, moving sqrt x to the right
side:

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

We'll raise to square both
sides:

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

We'll eliminate x both
sides:

sqrt(1-x) = 1 -
2sqrtx

We'll raise to square
again:

1 - x = 1 - 4sqrtx +
4x

We'll eliminate 1:

4sqrtx -
4x - x = 0

4sqrtx - 5x =
0

4sqrtx = 5x

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

16x =
25x^2

We'll subtract 16
x:

25x^2 - 16x = 0

We'll
factorize by x:

x(25x - 16) =
0

x1 = 0

25x =
16

x2 = 16/25

Since both
values of x are in the range of admissible values, we'll accept them as solutions of the
given equation.

x1 = 0 and x2 =
16/25.