Find $x-\sqrt{7x}$ given that $x - \sqrt{\frac7x}=8$

Question

If $ x - \sqrt{\frac{7}{x}}=8$ then $x-\sqrt{7x}=\text{?}$

I used some ways, but couldn't get the right form :) by the way, the answer is $1$.

Thanks in advance.

Answer 1

Let $x=7y^2,$ where $y>0$.

Thus, $$7y^2-\frac{1}{y}=8$$ or $$7y^3-8y-1=0$$ or $$7y^3+7y^2-7y^2-7y-y-1=0$$ or $$(y+1)(7y^2-7y-1)=0.$$ Thus, $$x-\sqrt{7x}=7y^2-7y=1.$$ Done!

Answer 2

\begin{eqnarray*} x-\sqrt{\frac{7}{x}}=8 \\ x-8 =\sqrt{\frac{7}{x}} \end{eqnarray*} Now square both sides and multiply by $x$ \begin{eqnarray*} x^2-16x+64=\frac{7}{x} \\ x^3-16x^2+64x-7=0 \\ (x-7)(x^2-9x+1)=0 \end{eqnarray*} Now assume $x$ does not equal $7$ and taking the positive square root will give \begin{eqnarray*} x^2-9x+1=0 \\ x^2-2x+1= 7x \\ (x-1)^2 =7x \\ x-1 = \sqrt{7x} \\ x-\sqrt{7x}= \color{red}{1}. \end{eqnarray*}