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

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(D) For the quadratic equation $x^2 + px + q = 0$,the sum of the roots is $-p$ and the product of the roots is $q$.Given roots are $	an 30^\circ$ and

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$3$

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(D) For the quadratic equation $x^2 + px + q = 0$,the sum of the roots is $-p$ and the product of the roots is $q$.Given roots are $	an 30^\circ$ and $	an 15^\circ$.So,$-p = 	an 30^\circ + 	an 15^\circ$ and $q = 	an 30^\circ \cdot 	an 15^\circ$.Using the identity $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have:$	an 45^\circ = \frac{	an 30^\circ + 	an 15^\circ}{1 - 	an 30^\circ 	an 15^\circ} = 1$.Substituting the values of $p$ and $q$:$1 = \frac{-p}{1 - q} \Rightarrow 1 - q = -p \Rightarrow p - q = -1$.We need to find the value of $2 + q - p$.Since $p - q = -1$,then $q - p = 1$.Therefore,$2 + q - p = 2 + 1 = 3$.

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

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