Question

Solve the following equation for x :
$\sqrt{x}+\sqrt{x-\sqrt{1-x}}=2$

Answer 1

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

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

$x\sqrt{1-x}=4+x-4\sqrt{x}$ (squaring)

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

$1-x=16(\sqrt{x}-1{)}^2$ (squaring)

$1-x=16(x+1-2\sqrt{x})$

$1-x=16x+16-32\sqrt{x}$

$\Rightarrow 15+17x=32\sqrt{x}$

$\Rightarrow 225+289x^2+513=1024x$ (squaring)

$\therefore 289x^2-514x+225=0$

$\therefore 289x^2-289x-225x+225=0$

$\therefore 289x(x-1)-225(x-1)=0$

$\Rightarrow (289x-225)(x-1)=0$

roots $\therefore x\ge 1$ So, $x\ne \frac{225}{289}$

$\therefore$ root : $x=1$