<s>[INST] Solve the equation $x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$ Solve the equation $$x+\dfrac{x}{\sqrt{x^2-1}}=\dfrac{35}{12}.$$
The equation is defined for $x\in\left(-\infty;-1\right)\cup\left(1;+\infty\right).$ Now I am thinking how to get rid of the radical in the denominator, but I can't come up with anything. Thank you!
 [/INST] From the equation solve for $\sqrt{x^2-1}$ and get
$$\sqrt{x^2-1} = \frac{x}{\frac{35}{12}-x}$$
now raise to the square and get an equation of degree $4$. Only $2$ of its roots will be roots of the initial equation, since they also include the   roots of $x-\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}$.
Worth plotting the graph of the function $x + \frac{x}{\sqrt{x^2-1}}$ on $(1, \infty)$. It it convex, with a minimum at $\sqrt{2}$, which is in between the roots $\frac{5}{4}$ and $\frac{5}{3}$ of our equation.
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