If the ratio of $HM$ between two positive numbers $a$ and $b$ ($a>b$) is to their $GM$ as $12$ to $13$, find $a:b$
My attempts:
$\dfrac{HM}{GM}=\dfrac{12}{13}\implies \dfrac{\dfrac{2ab}{a+b}}{\sqrt{ab}}=\dfrac{12}{13}\implies 13\sqrt{ab}=6(a+b)$
$\dfrac{13}{6}=\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b}{a}}$
Now, $6=2\cdot 3$ only (if not concern with $1$)
Hence I tried to break $\dfrac{13}{6}=\dfrac{c_1}{2}+\dfrac{c_2}{3}$
A bit more trial and error gives $c_1=3$ and $c_2=2$.
Hence ratio is $9:4$
But I want to know the method in which trials and errors are not incorporated, please help.
HINT:
Let $\sqrt{\dfrac ab}=p\implies p+\dfrac1p=\dfrac{13}6\iff0=6p^2-13p+6=(3p-2)(2p-3)$
If $p=\dfrac23,\dfrac ab=\dfrac49,\dfrac a4=\dfrac b9=k$(say)
$a=4k,b=?$