An algebra problem by Bala vidyadharan

{ x }^{ 2 }+\quad px\quad +\quad q\quad =\quad 0

Sum Of the Roots = \tan { (30) } +\quad \tan { (15) } \quad =\quad -p

Product Of The Roots = \tan { (30) } *\tan { (15) } \quad =\quad q

\therefore \quad 2\quad +\quad q\quad -\quad p\quad = \quad 2\quad +\quad \tan { (30) } *\tan { (15) } \quad +\quad \tan { (30) } +\quad \tan { (15) }

Simplify:

2\quad +\quad \frac { \sqrt { 3 } }{ 3 } *\frac { 2\quad -\quad \sqrt { 3 } }{ 1 } \quad +\quad \frac { \sqrt { 3 } }{ 3 } \quad +\quad \quad 2\quad -\quad \sqrt { 3 } \Rrightarrow \quad 3

Hence Answer : \boxed { 3 }

From

$\begin{aligned}x=&\ \tan{30^{\circ}}\quad&\text{and}\quad& \quad\quad\quad\quad\quad x&=&\ \tan{15^{\circ}}\\x=&\ \dfrac{\sqrt3}3\quad&\text{and}\quad& \quad\quad\quad\quad\quad x&=&\ 2-\sqrt3\\x-\dfrac{\sqrt3}3=&\ 0\quad&\text{and}\quad& x-(2-\sqrt3)&=&\ 0\end{aligned}$

$\begin{aligned}\left(x-\dfrac{\sqrt3}3\right)(x-(2-\sqrt3))=&\ 0\\x^2-(2-\sqrt3)x-\left(\dfrac{\sqrt3}3\right)x+\left(\dfrac{\sqrt3}3\right)(2-\sqrt3)=&\ 0\\x^2-\left(2-\sqrt3+\dfrac{\sqrt3}3\right)x+\left(\dfrac{2\sqrt3-3}3\right)=&\ 0\\x^2-\left(2-\left(\dfrac{3\sqrt3-\sqrt3}3\right)\right)x+\left(\dfrac{2\sqrt3}3-1\right)=&\ 0\\x^2-\left(2-\dfrac{2\sqrt3}3\right)x+\left(\dfrac{2\sqrt3}3-1\right)=&\ 0\\x^2+\left(\dfrac{2\sqrt3}3-2\right)x+\left(\dfrac{2\sqrt3}3-1\right)=&\ 0\end{aligned}$

Thus, $p=\dfrac{2\sqrt3}3-2$ and $q=\dfrac{2\sqrt3}3-1$

Hence,

$\begin{aligned}2+q-p=&\ 2+\left(\dfrac{2\sqrt3}3-1\right)-\left(\dfrac{2\sqrt3}3-2\right)\\=&\ 2+\dfrac{2\sqrt3}3-1-\dfrac{2\sqrt3}3+2=\boxed3\end{aligned}$