Question

The number of real roots of the equation $\sqrt{x}+\sqrt{x-\sqrt{1-x}}=1$ is

Answer 1

$\sqrt{x}+\sqrt{x-\sqrt{1-x}}=1\cdots \left(1\right)$
$\Rightarrow \sqrt{x-\sqrt{1-x}}=1-\sqrt{x}$
Squaring on both sides,
$\Rightarrow x-\sqrt{1-x}=1+x-2\sqrt{x}\Rightarrow -\sqrt{1-x}=1-2\sqrt{x}$
Again squaring on both sides,
$\Rightarrow 1-x=1+4x-4\sqrt{x}\Rightarrow 5x=4\sqrt{x}$
Again squaring on both sides,
$25x^2=16x\Rightarrow x(25x-16)=0\Rightarrow x=0,\frac{16}{25}$
Putting $x=0$ in the equation $\left(1\right)$,
$0+\sqrt{-1}=1$ which is not possible.
Now putting $x=\frac{16}{25}$
$\frac{4}{5}+\sqrt{\frac{16}{25}-\sqrt{\frac{9}{25}}}=\frac{4}{5}+\sqrt{\frac{1}{25}}=\frac{4}{5}+\frac{1}{5}=1$

Hence, only one possible solution.