<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] 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 $$\left(\dfrac{35}{12}\right)^2=u^2+2u\iff(u+1)^2=\left(\dfrac{37}{12}\right)^2\implies u=\dfrac{25}{12}\text{ as }u>0$$
$\sin2t=\dfrac2u=\dfrac{24}{25}\implies\cos2t=\pm\dfrac7{25}$
$x=\dfrac3{\cos t}=+\dfrac6{\sqrt{2(1+\cos2t)}}=?$
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