Solve $x > \sqrt{(1-x)}$.
Answer & explanation
Correct answer: option 1
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), $x∈\left(\frac{\sqrt{5}-1}{2},1\right]$