1. Given equations and root-sum sequences

We have: $x^2+\sqrt{3}x-16=0$ with roots $\alpha,\beta$, and $x^2+3x-1=0$ with roots $\gamma,\delta$.

Define $P_n=\alpha^n+\beta^n, \qquad Q_n=\gamma^n+\delta^n.$

We need to find $\frac{P_{25}+\sqrt{3}P_{24}}{2P_{23}}+\frac{Q_{25}-Q_{23}}{Q_{24}}.$

2. Recurrence relation for $P_n$

Since $\alpha,\beta$ satisfy $x^2+\sqrt{3}x-16=0,$ we have for each root $r$: $r^2=-\sqrt{3}r+16.$ Multiplying by $r^{n-2}$, $r^n=-\sqrt{3}r^{n-1}+16r^{n-2}.$ Adding for $r=\alpha,\beta$ gives: $P_n=-\sqrt{3}P_{n-1}+16P_{n-2}.$

Now compute the required numerator: $P_{25}+\sqrt{3}P_{24}.$ Using the recurrence for $n=25$, $P_{25}=-\sqrt{3}P_{24}+16P_{23}.$ Hence, $P_{25}+\sqrt{3}P_{24}=16P_{23}.$ Therefore, $\frac{P_{25}+\sqrt{3}P_{24}}{2P_{23}}=\frac{16P_{23}}{2P_{23}}=8.$

3. Recurrence relation for $Q_n$

Since $\gamma,\delta$ satisfy $x^2+3x-1=0,$ for each root $r$: $r^2=-3r+1.$ Multiplying by $r^{n-2}$, $r^n=-3r^{n-1}+r^{n-2}.$ Adding for $r=\gamma,\delta$ gives: $Q_n=-3Q_{n-1}+Q_{n-2}.$

Now for $n=25$, $Q_{25}=-3Q_{24}+Q_{23}.$ So, $Q_{25}-Q_{23}=-3Q_{24}.$ Therefore, $\frac{Q_{25}-Q_{23}}{Q_{24}}=-3.$

4. Add the two parts

Thus the required value is $8+(-3)=5.$

5. Check with options

The correct option is: $\boxed{5}$ which is Option C.

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.