Is there a better way to solve this equation? I came across this equation:
$x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4}$
Wolfram Alpha found 2 roots: $x=5$ and $x=\dfrac{15}{4}$, which "coincidentally" add up to $\dfrac{35}{4}$. So I'm thinking there should be a better way to solve it than the naïve way of bringing the fractions together and then squaring. Is there any?
 A: As $x>0$
WLOG $x=3\sec t, 0< t<\dfrac\pi2\implies \sin t,\cos t>0$
$$\dfrac{35}4=3\sec t+\dfrac{3\sec t}{3\tan t} \iff\sec t+\csc t=\dfrac{35}{12}$$
Method$\#1:$
Let $\sin t+\cos t=u$
$$\dfrac{35}{12}=\dfrac{2u}{u^2-1}$$
$$\iff0=35u^2-24u-35=7u(5u-7)+5(5u-7)=(5u-7)(7u+5)$$
As $u>0,  u=\sin t+\cos t=\dfrac75$
Now use $(\sin t+\cos t)^2+(\cos t-\sin t)^2=2$
$\cos t-\sin t=\pm\sqrt{2-\left(\dfrac75\right)^2}=\pm\dfrac15$
$2\cos t=\cos t-\sin t+\cos t+\sin t=\dfrac{7\pm1}5$
Method$\#2:$
As $\sec^2t+\csc^2t=\sec^2t\csc^2t$
Let $u=\sec t\csc t=\dfrac2{\sin2t}$
Squaring we get $$\left(\dfrac{35}{12}\right)^2=u^2+2u\iff(u+1)^2=\left(\dfrac{37}{12}\right)^2\implies u=\dfrac{25}{12}\text{ as }u>0$$
$\sin2t=\dfrac2u=\dfrac{24}{25}\implies\cos2t=\pm\dfrac7{25}$
$x=\dfrac3{\cos t}=+\dfrac6{\sqrt{2(1+\cos2t)}}=?$
A: We see that we need $x>3$ then let $x=\frac3{\cos y}$ with $y\in\left(0,\frac \pi 2\right)$
$$x + \dfrac{3x}{\sqrt{x^2 - 9}} = \dfrac{35}{4} \iff  \frac1{\cos y}+\frac1{\sin y}=\dfrac{35}{12}$$
and by half tangent identities by $t=\tan \frac y2$ we obtain
$$\frac{1+t^2}{1-t^2}+\frac{1+t^2}{2t}=\dfrac{35}{12} \iff (3t-1)(2t-1)(t^2-7t-6)=0$$
and $t=\frac 13, \frac12$ lead to the answer indeed


*

*$\dfrac3{\cos(2\cdot \arctan\left(\frac13\right)}=5$

*$\dfrac3{\cos(2\cdot \arctan\left(\frac12\right)}=\frac{15}4$
A: It' not a coincidence.   Let $f(x)=x + g(x)$ where $g(g(x))=x$, that is, $g$ is an involution.  We'll show that $f(x)=f(a-x)$, provided that $f(x)=a$.
Assume $f(x)=a$, that is $a-x=g(x)$. Now
$$f(a-x)=f(g(x))=g(x)+g(g(x))=g(x)+x=f(x)$$.
In our case define 
$$g(x)=\frac{3x}{\sqrt{x^2 - 9}}.$$
Indeed, $g$ is an involution as
$$g(g(x))=\dfrac{3\frac{3x}{\sqrt{x^2 - 9}}}{\sqrt{\frac{9x^2}{x^2 - 9} - 9}}=
\dfrac{x\frac{9}{\sqrt{x^2 - 9}}}{3\sqrt{\frac{x^2}{x^2 - 9} - 1}}=
\dfrac{x\frac{9}{\sqrt{x^2 - 9}}}{3\sqrt{\frac{9}{x^2 - 9}}}=x.
$$
A: Like Solve the equation $x^2+\frac{9x^2}{(x+3)^2}=27$
Clearly $x>0$
Let $\dfrac{3x}{\sqrt{x^2-9}}=y$ so that $x+y=\dfrac{35}4$
$$\implies9x^2=x^2y^2-9y^2\iff\dfrac19=\dfrac1{x^2}+\dfrac1{y^2}=\dfrac{(x+y)^2-2xy}{(xy)^2}$$
Use $x+y=\dfrac{35}4$ and $xy>0$ for find $xy=\dfrac{75}4$
So, $x,y$ are the roots of $$t^2-\dfrac{35}4t+\dfrac{75}4=0$$
