If the roots of the quadratic equation x^2 + | vedclass

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(D) Given the quadratic equation $x^2 + px + q = 0$ with roots $\alpha = 	an 30^\circ$ and $\beta = 	an 15^\circ$. From the relation between roots a

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$3$

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(D) Given the quadratic equation $x^2 + px + q = 0$ with roots $\alpha = 	an 30^\circ$ and $\beta = 	an 15^\circ$. From the relation between roots and coefficients,we have: Sum of roots: $\alpha + \beta = -p$ Product of roots: $\alpha \beta = q$ We know $	an 30^\circ = \frac{1}{\sqrt{3}}$ and $	an 15^\circ = 	an(45^\circ - 30^\circ) = \frac{	an 45^\circ - 	an 30^\circ}{1 + 	an 45^\circ 	an 30^\circ} = \frac{1 - 1/\sqrt{3}}{1 + 1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{4-2\sqrt{3}}{2} = 2 - \sqrt{3}$. Thus,$p = -(\alpha + \beta) = -(\frac{1}{\sqrt{3}} + 2 - \sqrt{3}) = -(\frac{1 + 2\sqrt{3} - 3}{\sqrt{3}}) = -(\frac{2\sqrt{3} - 2}{\sqrt{3}}) = \frac{2 - 2\sqrt{3}}{\sqrt{3}}$. And $q = \alpha \beta = \frac{1}{\sqrt{3}}(2 - \sqrt{3}) = \frac{2}{\sqrt{3}} - 1$. Now,calculate $2 + q - p$: $2 + (\frac{2}{\sqrt{3}} - 1) - (\frac{2 - 2\sqrt{3}}{\sqrt{3}}) = 1 + \frac{2}{\sqrt{3}} - \frac{2}{\sqrt{3}} + \frac{2\sqrt{3}}{\sqrt{3}} = 1 + 2 = 3$.

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

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