Evaluate $x+4/x$ when $x$ is a solution of $x-6/\sqrt{x}=11$ I have a question that I see in a math text.
I asked to my teacher but she couldn't find the solution, too. 
If 
$$x - \frac{6}{\sqrt{x}}=11,$$
then to what is equal
$$x + \frac{4}{x}\text{ ?}$$
 A: We can solve explicitly for $x$. Let $y=\sqrt{x}$. Easily we find that $y^3-11y-6=0$.
This has the obvious root $y=-3$. (In case I got lucky, I searched for integer roots. These must divide $6$, so the search was short.)
But $-3$ is no good for our purposes, since we are presumably looking for real $x$, and $-3$ is not the square of a real number.
For the other roots, divide $y^3-11y-6$ by the polynomial $y+3$. We get a quadratic. The roots can be found explicitly.  We now know $y$ and therefore by squaring, we know $x$. I got $x=\dfrac{13+3\sqrt{17}}{2}$ (the other root is negative). 
Now calculate $x+\frac{4}{x}$. A small miracle of cancellation happens, since $(13+3\sqrt{17})(13-3\sqrt{17})=16$. 
A: AS $x-\frac{6}{\sqrt{x}}-11=(\sqrt{x}+3)(\sqrt{x}-\frac{2}{\sqrt{x}}-3)$, so from $x-\frac{6}{\sqrt{x}}=11$ and $\sqrt{x}\geq0$, we can obtain that $\sqrt{x}-\frac{2}{\sqrt{x}}=3$, squaring at both sides, i.e. $(\sqrt{x}-\frac{2}{\sqrt{x}})^{2}=x+\frac{4}{x}-4=9$, so $x+\frac{4}{x}=13$.
A: Let $\sqrt{x}=t, t\ge0$, search t^3 - 11t - 6 in wolframalpha, we get the only positive solution $\sqrt{x} = t = \frac{1}{2}(3+\sqrt{17})$. Plug it in to $x+\frac{4}{x}$ we get the answer.
