I2.1 Determine the positive real root, $\alpha$, of $\sqrt{(x+\sqrt{x})}-\sqrt{(x-\sqrt{x})}=\sqrt{x}$.
$$
\begin{array}{l}
{[\sqrt{(x+\sqrt{x})}-\sqrt{(x-\sqrt{x})}]^{2}=x} \\
x+\sqrt{x}-2 \sqrt{x^{2}-x}+x-\sqrt{x}=x \\
x=2 \sqrt{x^{2}-x} \\
x^{2}=4\left(x^{2}-x\right) \\
3 x^{2}=4 x
\end{array}
$$
$x=0$ (rejected) or $\frac{4}{3}$
Check: When $x=\frac{4}{3}$,
$$
\begin{array}{l}
\text { L.H.S. }=\sqrt{\left(\frac{4}{3}+\sqrt{\frac{4}{3}}\right)}-\sqrt{\left(\frac{4}{3}-\sqrt{\frac{4}{3}}\right)}=\sqrt{\frac{4+2 \sqrt{3}}{3}}-\sqrt{\frac{4-2 \sqrt{3}}{3}}=\frac{\sqrt{3}+1}{\sqrt{3}}-\frac{\sqrt{3}-1}{\sqrt{3}}=\sqrt{\frac{4}{3}}=\text { R.H.S. } \\
\therefore \alpha=\frac{4}{3}
\end{array}
$$