Question

Solve:$\sqrt{x}+\sqrt{x+\sqrt{1-x}}=1$

Answer 1

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

Squaring both sides, we get

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

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

Squaring on both sides, we get

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

$\Rightarrow -5x=-4\sqrt{x}$

Squaring both sides, we get

$\Rightarrow 25x^2=16x$

$\Rightarrow x(25x-16)=0$

$\therefore x=0,\frac{16}{25}$

For $x=0,\sqrt{x}+\sqrt{x+\sqrt{1-x}}=0+\sqrt{0+\sqrt{1-0}}=1$

For $x=\frac{16}{25},$
$=\sqrt{\frac{16}{25}}+\sqrt{\frac{16}{25}+\sqrt{1-\frac{16}{25}}}$
$=\frac{4}{5}+\sqrt{\frac{16}{25}+\sqrt{\frac{25-16}{25}}}=\frac{4}{5}+\sqrt{\frac{16}{25}+\sqrt{\frac{9}{25}}}$
$=\frac{4}{5}+\sqrt{\frac{16}{25}+\frac{3}{5}}=\frac{4}{5}+\sqrt{\frac{16+15}{25}}=\frac{4}{5}+\frac{31}{25}\ne 1$

Hence $x=0$ is the only solution.