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!
After moving the $x$ to the right side and squaring, it's essentially the quartic equation $$144 x^4 - 840 x^3 + 937 x^2 + 840 x - 1225 = 0.$$
This can be factorized over $\Bbb Q$: $$(12x^2 - 35x - 49)(4x - 5)(3x - 5) = 0$$ and you easily get all four real roots from here.

EDIT:
I should confess that I didn't do the factorisation by hand. Instead, I used Sage, with the following code (which you can also try on the linked page):
R.&lt;x&gt; = ZZ[]
(144*x^4 - 840*x^3 + 937*x^2 + 840*x - 1225).factor()

and it gave the result in no time.
Complete the square.  Then solve to find $x = \dfrac54$ or $x=\dfrac53$.