<s>[INST] How to solve $\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$ The equation is $$\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$$
I cannot find any way to simplify this, other than squaring repeatedly, which is, btw, again not simplifying. Also tried substitutions like $1-x = t^2$ and trigonometric substitutions. None seem to work.
By trial and error, I find one root to be $\dfrac{16}{25}$. Is there a general way to treat this equation?
EDIT
The domain of definition of this function seems to be quite small, some subset of $(0,1)$. Also we know that there is only one root as the function is strictly increasing.
 [/INST] $$\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$$
$$\sqrt{x-\sqrt{1-x}} = 1 -\sqrt{x}$$
$$x-\sqrt{1-x} = (1 -\sqrt{x})^2$$
$$x-\sqrt{1-x} = x- 2\sqrt{x} +1$$
$$\sqrt{1-x} = 2\sqrt{x} +1$$
$$1-x = (2\sqrt{x} +1)^2$$
$$1-x = 1 + 4 \sqrt{x} + 4 x$$
$$- 5x= 4 \sqrt{x} $$
$$- \frac{5}{4}x= \sqrt{x} $$
$$\left(- \frac{5}{4}x\right)^2= x $$
$$\frac{25}{16}x^2= x $$
$$\frac{25}{16}= \frac{1}{x} $$
$$\frac{16}{25}= x $$
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