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 properties of roots,the s

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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 properties of roots,the sum of roots is $\alpha + \beta = -p$ and the product of roots is $\alpha \beta = q$.We know that $	an(30^\circ + 15^\circ) = 	an 45^\circ = 1$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have:$1 = \frac{-p}{1 - q}$.This implies $1 - q = -p$,or $p - q = -1$,which can be rewritten as $p + q = 1$ is incorrect; rather $1 - q = -p$ $\Rightarrow p - q = -1$ $\Rightarrow q - p = 1$.We need to find the value of $2 + q - p$.Substituting $q - p = 1$,we get $2 + (1) = 3$.

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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