Radicals: Solve for x from $\sqrt{x - \sqrt{1 - x}} = 1 - \sqrt{x}$

$\sqrt{x - \sqrt{1 - x}} = 1 - \sqrt{x}$

$x - \sqrt{1 - x} = \left( 1 - \sqrt{x} \right)^2$

$x - \sqrt{1 - x} = 1 - 2\sqrt{x} + x$

$\sqrt{1 - x} = 2\sqrt{x} - 1$

$1 - x = \left( 2\sqrt{x} - 1 \right)^2$

$1 - x = 4x - 4\sqrt{x} + 1$

$4\sqrt{x} = 5x$

$16x = 25x^2$

$16 = 25x$

$x = \dfrac{16}{25}$

Pwde po ba idirect sa calcu?