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!
 A: One can use trigonometric functions to avoid squaring both sides to get a degree 4 polynomial. Clearly $x>1$. Let $x=\sec t$, $t\in(0,\frac{\pi}{2})$. Then the equation becomes
$$ \sec t+\csc t=\frac{35}{12} $$
or
$$ \frac{\sin t+\cos t}{\sin t\cos t}=\frac{35}{12}. $$
Squaring both sides gives
$$ \frac{1+\sin(2t)}{\sin^2(2t)}=\frac{35^2}{24^2} $$
which is equivalent to
$$ 35^2\sin^2(2t)-24^2\sin(2t)-24^2=0 $$
or
$$ (25\sin(2t)-24)(49\sin(2t)+24)=0. $$
Since $\sin(2t)>0$, one has
$$ \sin(2t)=\frac{24}{25} $$
which gives
$$ \cos(2t)=\pm\frac{7}{25}. $$
So
$$ \cos t=\sqrt{\frac{1+\cos(2t)}{2}}=\frac{3}{5} \text{ or } \frac{4}{5} $$
and hence
$$ x=\frac{5}{3} \text{ or }\frac{5}{4}. $$
A: 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.<x> = ZZ[]
(144*x^4 - 840*x^3 + 937*x^2 + 840*x - 1225).factor()

and it gave the result in no time.
A: Subtract $x$ from both sides, square it, and then multiply both sides by $x^2-1$.  You’ll get a quartic equation in $x$.
A: Let $x=5/u$ with $0\lt u\lt5$ (there are no solutions with $x\lt0$ since $35/12\gt0$). The equation becomes
$${1\over u}+{1\over\sqrt{25-u^2}}={7\over12}={1\over3}+{1\over4}$$
The solutions $u=3$ and $u=4$ (corresponding to $x=5/3$ and $x=5/4$) are easy to see by inspection.  To show there are no other solutions, it suffices to note that
$$\begin{align}
f(u)=u^{-1}+(25-u^2)^{-1/2}
&\implies f'(u)=-u^{-2}+u(25-u^2)^{-3/2}\\
&\implies f''(u)=2u^{-3}+(25-u^2)^{-3/2}+3u^2(25-u^2)^{-5/2}\gt0
\end{align}$$
so the curve is convex on $(0,5)$ and thus $f(u)$ cannot take any value more than twice.
A: Hint:
Multiply for $12\sqrt{x^2-1}$ your initial equation,
you have $$12(\sqrt{x^2-1})x+12x=35\sqrt{x^2-1} \iff 12x(\sqrt{x^2-1}+1)=35\sqrt{x^2-1}$$
and square (two times to delete the sqrt) to the left and the right.
$$144x^2(\sqrt{x^2-1}+1)^2=35^2(x^2-1)$$
$$144x^2(x^2-1+1+2\sqrt{x^2-1})=35^2(x^2-1)$$
A: 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.
A: 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.)
A: Complete the square.  Then solve to find $x = \dfrac54$ or $x=\dfrac53$.

