let-and-be-the-root-of-the-equation-ax-2-bx-c-0-find-the-equation-whose-roots-are-1-1-and-1-1-

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