If the roots of the quadratic equation $$x^2 + px + q = 0$$ are $$\tan 30^\circ$$ and $$\tan 15^\circ$$, respectively then the value of $$2 + q - p$$ is

Let the roots of the quadratic $$x^{2}+px+q=0$$ be $$r_{1}$$ and $$r_{2}$$.
Given: $$r_{1}=\tan 30^\circ$$ and $$r_{2}=\tan 15^\circ$$.

Step 1: Evaluate the numerical values of the roots.

For $$30^\circ$$ we know $$\tan 30^\circ = \frac{1}{\sqrt3}$$, so $$r_{1}=\frac{1}{\sqrt3}$$.

To get $$\tan 15^\circ$$, use the formula for $$\tan(A-B)$$:
$$\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}.$$
Choose $$A=45^\circ,\, B=30^\circ$$; thus

$$\tan 15^\circ = \tan(45^\circ-30^\circ) =\frac{\tan45^\circ-\tan30^\circ}{1+\tan45^\circ\tan30^\circ} =\frac{1-\frac{1}{\sqrt3}}{1+1\cdot\frac{1}{\sqrt3}}$$

Simplify numerator and denominator by multiplying top and bottom by $$\sqrt3$$:

$$\tan15^\circ =\frac{\sqrt3-1}{\sqrt3+1} =2-\sqrt3.$$
Hence $$r_{2}=2-\sqrt3$$.

Step 2: Relate coefficients to the roots.

For the quadratic $$x^{2}+px+q=0$$ with roots $$r_{1},r_{2}$$:
Sum of roots: $$r_{1}+r_{2}=-p.$$
Product of roots: $$r_{1}r_{2}=q.$$

Step 3: Compute $$r_{1}+r_{2}$$ and $$r_{1}r_{2}$$.

Sum:
$$r_{1}+r_{2} =\frac{1}{\sqrt3}+2-\sqrt3 =\frac{\sqrt3}{3}+2-\sqrt3 =2-\frac{2\sqrt3}{3}.$$

Product:
$$r_{1}r_{2} =\frac{1}{\sqrt3}\,(2-\sqrt3) =\frac{2-\sqrt3}{\sqrt3} =\frac{2}{\sqrt3}-1.$$

Thus
$$p=-(r_{1}+r_{2})= -\left(2-\frac{2\sqrt3}{3}\right),$$
$$q=r_{1}r_{2}= \frac{2}{\sqrt3}-1.$$

Step 4: Evaluate $$2+q-p$$.

First rewrite $$p$$ with a positive denominator: $$p=-2+\frac{2\sqrt3}{3}.$$

Now calculate:

$$$ \begin{aligned} 2+q-p &=2+\left(\frac{2}{\sqrt3}-1\right)-\left(-2+\frac{2\sqrt3}{3}\right)\\[4pt] &=2-1+2+\frac{2}{\sqrt3}-\frac{2\sqrt3}{3}\\[4pt] &=3+\left(\frac{2}{\sqrt3}-\frac{2\sqrt3}{3}\right). \end{aligned} $$$

Notice that $$\frac{2}{\sqrt3} = \frac{2\sqrt3}{3},$$ so the two radical terms cancel:

$$\frac{2}{\sqrt3}-\frac{2\sqrt3}{3}=0.$$

Therefore

$$2+q-p = 3.$$

Hence the required value is 3.
Option B which is: 3