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

Question

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?

Answer 1

Starting with what you have written we quickly see that $$0=x^3-18x^2+81x-8=(x-8)(x^2-10x+1)$$

It is easy to see that $x\neq 8$ so we must have $\boxed {x^2-10x+1=0}$

Now, go back to the original equation. Multiply by $x$ to get $$x^2-\sqrt {8x}=9x\implies x-\sqrt {8x}=10x-x^2$$

But the boxed equation tells us that $10x-x^2=1$ so $$\boxed {x-\sqrt {8x}=1}$$

Answer 2

Let $x=t^2,$ where $t>0$.

Thus, $$t^3-9t-\sqrt8=0$$ or $$t^3+\sqrt8t^2-\sqrt8t^2-8t-t-\sqrt8=0$$ or $$(t+\sqrt8)(t^2-\sqrt8t-1)=0$$ or $$x-\sqrt{8x}=1.$$

Answer 3

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