Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to? 
Given $x-\sqrt {\dfrac {8}{x}}=9$, what is $x-\sqrt{8x}$ equal to?

My attempt:
We have
\begin{align}
x-\sqrt {\dfrac {8}{x}}=9
\implies -\sqrt {\dfrac {8}{x}}=9-x
\implies \dfrac {8}{x}=(9-x)^2
\end{align}
How can I proceed?

$x-\dfrac {\sqrt {8x}}{x}=9,\Rightarrow x^{2}-\sqrt {8x}=9x$
$\begin{aligned}x^{2}-9x=\sqrt {8x}\\ x^{2}-8x=\sqrt {8x}+x\\ \left( x-\sqrt {8x}\right) \left( x+\sqrt {8x}\right) =\sqrt {8x}+x\\ x -\sqrt {8x}=1\end{aligned}$
