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!

Note that $x>1$. To avoid squaring equations, factorize directly as follows
\begin{align}
0 &=x+\frac{x}{\sqrt{x^2-1}}-\frac{35}{12}\\
&= \left(x-\frac{35}{12} + \frac{25}{12x}\right)
 + \left( \frac{x}{\sqrt{x^2-1}} -\frac{25}{12x}\right)\\
&=\frac1x \left( x^2-\frac{35}{12} x+ \frac{25}{12}\right)
 + \frac{x^2-\frac{25}{12}\sqrt{x^2-1}}{x\sqrt{x^2-1}}\\
& =\frac1x \left( x^2-\frac{35}{12} x+ \frac{25}{12}\right)
 + \frac{x^4-\frac{25^2}{12^2}({x^2-1})}{x\sqrt{x^2-1}\left( x^2+\frac{25}{12}\sqrt{x^2-1}\right)}\\
& =\frac1x \left( x-\frac{5}{4} \right)\left(x- \frac{5}{3}\right)
 + \frac{(x^2-\frac{25}{16})(x^2-\frac{25}{9} )}{x\sqrt{x^2-1}\left( x^2+\frac{25}{12}\sqrt{x^2-1}\right)}\\
& =\frac1x \left( x-\frac{5}{4} \right)\left(x- \frac{5}{3}\right)
\left( 1+ \frac{(x+\frac{5}{4})(x+\frac{5}{3} )}{\sqrt{x^2-1}\left( x^2+\frac{25}{12}\sqrt{x^2-1}\right)}\right)\\
\end{align}
which leads to the solutions $x= \frac54,\>\frac53$. (The last factor is always positive.)
