The roots of the equation 2x^2 − x + 3 = 0 are α and β if the roots of 3x^2 + px + q=0 are α + (1/α) and β + (1/(β )) find the value of p and q.


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Question Number 38012 by Rio Mike last updated on 20/Jun/18

$T h e r o o t s o f t h e e q u a t i o n$ $2 x^{2} - x + 3 = 0 a r e α a n d β$ $i f t h e r o o t s o f 3 x^{2} + p x + q = 0$ $a r e α + \frac{1}{α} a n d β + \frac{1}{β} f i n d t h e v a l u e$ $o f p a n d q .$
Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jun/18

$s p e n d t i m e t o r e a d b o o k . . . d e a d d i c t f r o m$ $v i r t u a l s c h o o l . . . .$
	Answered by MJS last updated on 20/Jun/18

	$2 x^{2} - x + 3 = 0$ $x = \frac{1}{4} \pm \sqrt{\frac{1}{16} - \frac{3}{2}} = \frac{1}{4} \pm \frac{\sqrt{23}}{4} i$ $α = \frac{1}{4} - \frac{\sqrt{23}}{4} i \Rightarrow \frac{1}{α} = \frac{1}{6} + \frac{\sqrt{23}}{6} i \Rightarrow α + \frac{1}{α} = \frac{5}{12} - \frac{\sqrt{23}}{12} i$ $β = \frac{1}{4} + \frac{\sqrt{23}}{4} i \Rightarrow \frac{1}{β} = \frac{1}{6} - \frac{\sqrt{23}}{6} i \Rightarrow β + \frac{1}{β} = \frac{5}{12} + \frac{\sqrt{23}}{12} i$ $3 (x - \frac{5}{12} + \frac{\sqrt{23}}{12} i) (x - \frac{5}{12} - \frac{\sqrt{23}}{12} i) =$ $= 3 x^{2} - \frac{5}{2} x + 1$ $p = - \frac{5}{2}$ $q = 1$
	Answered by ajfour last updated on 20/Jun/18

	$α β = \frac{3}{2}, α + β = \frac{1}{2}$ $(α + \frac{1}{α}) (β + \frac{1}{β}) = \frac{q}{3}$ $α + β + \frac{1}{α} + \frac{1}{β} = - \frac{p}{3}$ $. . . . . . . . .$ $\Rightarrow α β + \frac{α}{β} + \frac{β}{α} + \frac{1}{α β} = \frac{q}{3}$ $(α + β) + \frac{α + β}{α β} = - \frac{p}{3}$ $\frac{3}{2} + \frac{{(\frac{1}{2})}^{2} - 2 (\frac{3}{2})}{\frac{3}{2}} + \frac{2}{3} = \frac{q}{3}$ $a n d \frac{1}{2} + \frac{(\frac{1}{2})}{(\frac{3}{2})} = - \frac{p}{3}$ $q = 3 [\frac{3}{2} - \frac{11}{6} + \frac{2}{3}] = 1$ $p = - 3 [\frac{1}{2} + \frac{1}{3}] = - \frac{5}{2} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jun/18

	$α + β = \frac{1}{2} α β = \frac{3}{2}$ $α + \frac{1}{α} + β + \frac{1}{β} = \frac{- p}{3}$ $α + β + \frac{α + β}{α β} = \frac{- p}{3}$ $\frac{1}{2} + \frac{1}{2} \times \frac{2}{3} = - \frac{p}{3}$ $\frac{5}{6} = - \frac{p}{3}$ $p = - \frac{5}{2}$ $(α + \frac{1}{α}) \times (β + \frac{1}{β}) = \frac{q}{3}$ $α β + \frac{α}{β} + \frac{β}{α} + \frac{1}{α β} = \frac{q}{3}$ $α β + \frac{1}{α β} + \frac{α^{2} + β^{2}}{α β} = \frac{q}{3}$ $\frac{3}{2} + \frac{2}{3} + \frac{{(α + β)}^{2} - 2 α β}{α β} = \frac{q}{3}$ $\frac{13}{6} + \frac{\frac{1}{4} - 2 \times \frac{3}{2}}{\frac{3}{2}} = \frac{q}{3}$ $\frac{13}{6} + \frac{\frac{1 - 12}{4}}{\frac{3}{2}} = \frac{q}{3}$ $\frac{13}{6} - \frac{11 \times 2}{12} = \frac{q}{3}$ $\frac{13 - 11}{6} = \frac{q}{3}$ $q = 1$
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