Recurrence which cannot be analyzed by master theorem
Suppose for simplicity that $n$ is a power of 2. Construct the tree corresponding to the recursion – each vertex is labelled by the current value of $n$, has three children, one of which is a leaf. We assume for now that the tree continues forever. Associate with each edge a weight, which is $1$, $a$, or $1$, depending on which term it corresponds to. The weight of a vertex is the product of all weights in the path from the root to the vertex.

Consider the generating series $$P(x) = \frac{1}{1-x-ax^2} = \sum_{m=0}^\infty (x+ax^2)^m.$$
The coefficient of $x^k$ in $P(x)$ is the total weight of vertices whose label is $n/2^k$. Therefore $\tilde T(n) = n^2 P(1/4)$, where the tilde denotes that there is no base case and the recursion continues forever (we will discuss this later on). Thus $\tilde T(n) = \omega(n^2)$ iff the power series $P(x)$ diverges at $x = 1/4$. This happens if $1/4 + a/4^2 \geq 1$, that is, if $a \geq 12$.

There is a slight problem with this argument, namely the tree doesn't continue forever. However, it is easy to see that all leaves are at depth $\Theta(\log n)$, or more accurately, at depths between $\log_4 n$ and $\log_2 n$ (assuming a base case $T(1) = 1$). This shows that
$$
\sum_{m=0}^{\log_4 n} \left(\frac{1}{4}+\frac{a}{16}\right)^m \leq T(n) \leq
\sum_{m=0}^{\log_2 n} \left(\frac{1}{4}+\frac{a}{16}\right)^m.
$$
When $a &lt; 12$, the preceding argument already shows that $T(n) = \Theta(n^2)$. When $a = 12$, these bounds show that $T(n) = \Theta(n^2 \log n)$. When $a &gt; 12$ these bounds are not enough to determine the asymptotics, though definitely $T(n) = \omega(n^2)$.

The Akra–Bazzi method is a general form of this argument. To solve your recursion, you first have to find the value of $p$ such that $(1/2)^p + a(1/4)^p = 1$. Given that, the Akra–Bazzi theorem tells you that
$$
T(n) = \Theta(n^p) \cdot \left(1 + \int_1^n \frac{u^2 \, du}{u^{p+1}}\right) = \begin{cases} \Theta(n^2) &amp; p &lt; 2, \\ \Theta(n^2\log n) &amp; p = 2, \\ \Theta(n^p) &amp; p &gt; 2. \end{cases}
$$
(I leave the case analysis needed to prove the formula on the right-hand side to you. It is very instructive.)
The critical point is when $p = 2$, and this happens when $1/4 + a/16 = 1$, that is, when $a = 12$, and so in this case $T(n) = \Theta(n^2 \log n)$. When $a &lt; 12$ we have $p &lt; 2$, and so in this case $T(n) = \Theta(n^2)$. These results agree with the calculations above. Finally, when $a &gt; 12$ we have $p &gt; 2$, and so in this case $T(n) = \Theta(n^p)$, the estimate which was too difficult for our crude method above.