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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 \dots 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 \dots .

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}

\dots \dots \dots

\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