Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

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

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

$\Rightarrow {\left(\sqrt{x-\sqrt{1-x}}\right)}^2={(1-\sqrt{x})}^2$ by squaring on both sides

$\Rightarrow x-\sqrt{1-x}=1+x-2\sqrt{x}$ by cancelling the same terms on both sides

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

$\Rightarrow {(-\sqrt{1-x})}^2={(1-2\sqrt{x})}^2$ by squaring on both sides

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

$\Rightarrow -x-4x=-4\sqrt{x}$ on simplification

$\Rightarrow {(-5x)}^2={(-4\sqrt{x})}^2$

$\Rightarrow 25x^2=16x$

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

$\Rightarrow x=0,25x-16=0$

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

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