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!
On
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.
$$144x^2(\sqrt{x^2-1}+1)^2=35^2(x^2-1)$$
$$144x^2(x^2-1+1+2\sqrt{x^2-1})=35^2(x^2-1)$$
On
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.
On
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.
On
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 $$ x=\frac{5}{3} \text{ or }\frac{5}{4}. $$
On
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.
On
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.)
Subtract $x$ from both sides, square it, and then multiply both sides by $x^2-1$. You’ll get a quartic equation in $x$.