<s>[INST] 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?
 [/INST] 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


*

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

*$\dfrac3{\cos(2\cdot \arctan\left(\frac12\right)}=\frac{15}4$
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