<s>[INST] Time Complexity in $\theta$ Notation $$T(n) = 2T\left(\frac{n}{2}\right) + T\left(\frac{n}{4}\right) + 5$$
What is the time complexity of the given algorithm in $\theta$ notation.
Thanks in advance.
 [/INST] First of all note that this recurrence only depends on the number of bits of $n$ and not on their values, which simplifies things considerably.
Suppose we wish to solve
$$T(n) = 2T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + 5$$
where we set $T(0)=0.$
We unroll the recursion to obtain an exact formula for all $n:$
$$T(n) = 5 \times \sum_{j=0}^{\lfloor \log_2 n \rfloor} 
[z^j] \frac{1}{1-2z-z^2}.$$
Now the roots of $1-2z-z^2 = 2-(z+1)^2$ are at $$-1 \pm \sqrt{2}$$
and their inverses are
$$\rho_{1,2} = 1 \pm \sqrt{2}.$$
It follows that
$$[z^j]   \frac{1}{1-2z-z^2} = c_1 (1+\sqrt{2})^j + c_2 (1-\sqrt{2})^j.$$
Once we determine the coefficients we get
$$[z^j]   \frac{1}{1-2z-z^2} = 
\frac{2+\sqrt{2}}{4} (1+\sqrt{2})^j+\frac{2-\sqrt{2}}{4} (1-\sqrt{2})^j.$$
The conclusion is that the closed and exact formula for $T(n)$ is
$$5\times 
\frac{2+\sqrt{2}}{\sqrt{2}\times 4} ((1+\sqrt{2})^{\lfloor \log_2 n \rfloor+1}-1)-
5\times
\frac{2-\sqrt{2}}{\sqrt{2}\times 4} ((1-\sqrt{2})^{\lfloor \log_2 n \rfloor+1}-1).$$
If desired we may re-write this as
$$-\frac{5}{2}+
5\times 
\frac{2+\sqrt{2}}{\sqrt{2}\times 4} (1+\sqrt{2})^{\lfloor \log_2 n \rfloor+1}-
5\times
\frac{2-\sqrt{2}}{\sqrt{2}\times 4} (1-\sqrt{2})^{\lfloor \log_2 n \rfloor+1}.$$
It follows that the complexity from the dominant asymptotic term is
$$\Theta\left((1+\sqrt{2})^{\lfloor \log_2 n \rfloor}\right)
= \Theta\left(2^{\log_2(1+\sqrt{2})\times \log_2 n}\right)
= \Theta\left(n^{\log_2(1+\sqrt{2})}\right).$$
The exponent here on $n$ is $$ 1.271553303,$$confirming the analysis by @Did.
A similar Master Theorem calculation can be found at this MSE link.
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