# 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

$$x + \frac{4}{x}\text{ ?}$$

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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 $x+\frac{4}{x}=13$.

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Thanks,the answers also say 13 :) –  Caner Korkmaz Oct 4 '12 at 17:47

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 $(13+3\sqrt{17})(13-3\sqrt{17})=16$.

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Once you have the quadratic $y^2-3y-2$, don't calculate the roots. Instead observe that $x+\frac4x=(y+\frac{-2}y)^2+4$. If $y$ is one root of the quadratic, tehn $\frac{-2}y$ is the other (from constant term$=-2$), hence the parentheses is the sum of the roots, which is 3 (from linear coefficient = $-3$). –  Hagen von Eitzen Oct 4 '12 at 15:18
Yes, I was operating mechanically, since it was obviously all over. –  André Nicolas Oct 4 '12 at 15:21
False answer, the answers if text says 13 –  Caner Korkmaz Oct 4 '12 at 17:45
I didn't give an answer, I gave $x$, from which the calculation of $x+\frac{4}{x}$ is not hard. If you calculate $x+\frac{4}{x}$, you will get $13$, because $x=\frac{13+3\sqrt{17}}{2}$ and after rationalizing the denominator using the last line of my answer you will get $\frac{4}{x}=\frac{13-3\sqrt{17}}{2}$. I had left this final computation to you, also the division of the cubicin $y$ by $y+3$. –  André Nicolas Oct 4 '12 at 18:08

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.

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