How to solve $x - 3\sqrt{\frac{5}{x}} = 8$ for $x - \sqrt{5x}$? I have a problem from my textbook. By using $x - 3\sqrt{\frac{5}{x}} = 8$ how can we find the value of $x - \sqrt{5x}$. I have derived the equation that's given so much, but i couldn't find the answer. Solvings or hints are appreciated.
 A: we can write $$x-8=3\sqrt{\frac{5}{x}}$$ after squaring this equation we get
$$x^2-16x+64=9\cdot \frac{5}{x}$$ multiplying by $$x\ne 0$$ we obtain
$$x^3-16x^2+64x-45=0$$ factorizing this equation we get
$$(x-5)(x^2-11x+9)=0$$ can you finish now?
A: The function $f(x)=x-3\sqrt{5\over x}$ increases from $-\infty$ at $x=0$ to $+\infty$ as $x\to\infty$, so the equation $x-3\sqrt{5\over x}=8$ has exactly one solution for $x\gt0$.  Multiplying through by $x$ and moving everything to the left hand side, we have
$$x^2-8x-3\sqrt{5x}=0$$
Now let $u=x-\sqrt{5x}$.  We can rewrite this equation as
$$x^2-11x+3u=0$$
On the other hand, moving the $-3\sqrt{5x}$ to the right hand side and squaring gives $x^4-16x^3+64x^2=45x$, or
$$(x^3-16x^2+64x-45)x=(x-5)(x^2-11x+9)x=0$$
But $x=0$ is obviously not a solution to $x-3\sqrt{5\over x}=8$.  Nor is $x=5$.  So the (unique positive) solution must satisfy $x^2-11x+9=0$.  But since it also satisfies $x^2-11x+3u=0$, we see that $9=3u$, or $u=3$.  We thus find that $x-\sqrt{5x}=3$.
