Solve $x > \sqrt{(1-x)}$.

$\left(\frac{\sqrt{5}-1}{2},1\right]$

Given inequality can be solved by squaring both sides. But sometimes squaring gives extraneous solutions which do not satisfy the original inequality. Before squaring we must restrict x for which terms in the given inequality are well defined.

$x > \sqrt{(1-x)}$. Here x must be positive.

Now, $\sqrt{1-x}$ is defined only when $1 - x ≥ 0$ or $x ≤ 1$

Thus $0 ≤ x ≤ 1$

Squaring given inequality both sides, $x^2>1-x$

$⇒x^2+x-1>0$

$⇒\left(x-\frac{-1-\sqrt{5}}{2}\right)\left(x-\frac{-1+\sqrt{5}}{2}\right)>0$

$⇒x<\frac{-1-\sqrt{5}}{2}$ or $x>\frac{-1+\sqrt{5}}{2}$

From (1) and (2), $x∈\left(\frac{\sqrt{5}-1}{2},1\right]$