let α and β be the root of the equation ax^2 +bx+c=0. find the equation whose roots are ((1/α)+(1/β)) and ((1/α)−(1/β))


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Question Number 172024 by Mikenice last updated on 23/Jun/22

$l e t α a n d β b e t h e r o o t o f t h e e q u a t i o n$ ${a x}^{2} + b x + c = 0 . f i n d t h e e q u a t i o n w h o s e r o o t s$ $a r e (\frac{1}{α} + \frac{1}{β}) a n d (\frac{1}{α} - \frac{1}{β})$
	Answered by Rasheed.Sindhi last updated on 23/Jun/22

	$G i v e n e q u a t i o n :$ ${a x}^{2} + b x + c = 0$ $α + β = - \frac{b}{a}, α β = \frac{c}{a}, α - β = \frac{\sqrt{b^{2} - 4 a c}}{a}$ $R e q u i r e d e q u a t i o n :$ $S u m o f t h e r o o t s = (\frac{1}{α} + \frac{1}{β}) + (\frac{1}{α} - \frac{1}{β})$ $= \frac{α + β}{α β} + \frac{- (α - β)}{α β}$ $= \frac{(- b / a) - (\sqrt{b^{2} - 4 a c}) / a}{(c / a)} = \frac{- b - \sqrt{b^{2} - 4 a c}}{c}$ $P r o d u c t o f r o o t s = (\frac{1}{α} + \frac{1}{β}) (\frac{1}{α} - \frac{1}{β})$ $= \frac{1}{α^{2}} - \frac{1}{β^{2}} = \frac{- (α + β) (α - β)}{{(α β)}^{2}}$ $= \frac{- (- b / a) (- b - \sqrt{b^{2} - 4 a c} / c)}{{(c / a)}^{2}}$ $= - \frac{a b}{c^{3}} (b + \sqrt{b^{2} - 4 a c})$ $x^{2} - (S u m o f t h e r o o t s) x + (p r o d u c t o f t h e r o o t s) = 0$ $x^{2} - (\frac{- b - \sqrt{b^{2} - 4 a c}}{c}) x - \frac{a b}{c^{3}} (b + \sqrt{b^{2} - 4 a c}) = 0$ $c^{3} x^{2} + c^{2} (b + \sqrt{b^{2} - 4 a c}) x - a b (b + \sqrt{b^{2} - 4 a c}) = 0$
	Commented by Mikenice last updated on 23/Jun/22

	$t h a n k s s i r$
	Commented by peter frank last updated on 23/Jun/22

	$thanks$
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