unorthodox solution of a special case of the master theorem

A hands-on approach is to find some hereditary upper and lower bounds of $T(n)$ by multiples of powers of $n$.

If $T(n)=T(n/2)+3T(n/4)+n$, one first centers the recursion using $S(n)=T(n)+cn$, then $$ S(n)=S(n/2)-cn/2+3T(n/4)-3cn/4+n+cn=S(n/2)+3S(n/4), $$ if $c=4$. From now on, $S(n)=T(n)+4n$.

Thus, the centering step is based on the affine part of the recursion, here $+n$.

Second, assume that $S(k)\bowtie c k^a$ for $k=n/2$ and $k=3n/4$ with $a\gt1$, where $\bowtie$ is either $\leqslant$ or $\geqslant$. Then $S(n)\bowtie cn^a/2^a+c3n^a/4^a=(1/2^a+3/4^a)cn^a$. From now on, assume that $a$ solves $$ \frac1{2^a}+\frac3{4^a}=1. $$ Then the property $S(n)\bowtie cn^a$ is hereditary for every fixed $c$.

Thus, the power step is based on the linear part of the recursion, here $T(n/2)+3T(n/4)$.

In particular, for some small enough $c$ and some large enough $C$, it holds that $$ cn^a-4n\leqslant T(n)\leqslant Cn^a-4n, $$ for every $n$. Since $a\gt1$ in the present case, the contribution of the linear part of the recursion wins hence all this is more than enough to prove that $$ T(n)=\Theta(n^a). $$ One thing is clear though, which is that the fact that $1/2$ and $1/4$ are inverses of powers of a unique prime, invoked in the OP, does not enter the picture. Nevertheless, to relate $a$ to the exponent $\rho_1$ in the OP, assume that $a=1+\log_2\rho$, then $$ \frac1{2^a}+\frac3{4^a}=\frac1{2\rho}+\frac3{4\rho^2}, $$ hence $\rho$ solves $4\rho^2=2\rho+3$, that is, $\rho=\frac14(1\pm\sqrt{13})$, as claimed in the OP.

Edit: About the paragraph justifying the bounty, I would check the obvious references, namely generatingfunctionology and Flajolet's books.

Look at the Akra-Bazzi theorem, it gives bounds for recurrences like yours. The proof is a blizzard of integrals...