If p and q are roots of 6x^2 + 10x + 1 = 0,th | vedclass

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(D) Given the quadratic equation $6x^2 + 10x + 1 = 0$.By Vieta's formulas,the sum of roots $p + q = -\frac{10}{6} = -\frac{5}{3}$ and the product of r

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$-1$

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(D) Given the quadratic equation $6x^2 + 10x + 1 = 0$.By Vieta's formulas,the sum of roots $p + q = -\frac{10}{6} = -\frac{5}{3}$ and the product of roots $pq = \frac{1}{6}$.Since $p+q  0$,both roots $p$ and $q$ are negative.Let $p = -a$ and $q = -b$,where $a, b > 0$.Then $	an^{-1} p + 	an^{-1} q = -(	an^{-1} a + 	an^{-1} b)$.Using the formula $	an^{-1} a + 	an^{-1} b = 	an^{-1} \left( \frac{a+b}{1-ab} ight)$ for $ab $	an^{-1} p + 	an^{-1} q = -	an^{-1} \left( \frac{p+q}{1-pq} ight) = -	an^{-1} \left( \frac{-5/3}{1-1/6} ight) = -	an^{-1} \left( \frac{-5/3}{5/6} ight) = -	an^{-1} (-2) = 	an^{-1} 2$.Wait,re-evaluating: $	an^{-1} p + 	an^{-1} q = -	an^{-1} \left( \frac{a+b}{1-ab} ight)$ where $a+b = 5/3$ and $ab = 1/6$.$	an^{-1} p + 	an^{-1} q = -	an^{-1} \left( \frac{5/3}{1-1/6} ight) = -	an^{-1} \left( \frac{5/3}{5/6} ight) = -	an^{-1} (2)$.Since $	an^{-1} (1) Thus,$-\frac{\pi}{3} Since $-1.047 < -	an^{-1} 2 < -0.785$,the greatest integer $[-	an^{-1} 2]$ is $-1$.

If $p$ and $q$ are roots of $6x^2 + 10x + 1 = 0$,then the value of $[\tan^{-1} p + \tan^{-1} q]$ is: {where $[x]$ denotes the greatest integer less than or equal to $x$}

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