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Question Number 63154 by Rio Michael last updated on 29/Jun/19

G i v e n t h a t α a n d β a r e t h e r o o t s o c t h e e q u a t i o n {a x}^{2} + b x + c = 0

. S h o w t h a t λ μ b^{2} = a c {(λ + μ)}^{2}, w h e r e \frac{α}{β} = \frac{λ}{μ} .

Commented by Prithwish sen last updated on 30/Jun/19

α + β = - \frac{b}{a} and α β = \frac{c}{a}

Now

{(α + β)}^{2} = \frac{α β {(α + β)}^{2}}{α β} = α β (\frac{α}{β} + \frac{β}{α} + 2)

\frac{b^{2}}{a^{2}} = \frac{c}{a} (\frac{λ}{μ} + \frac{μ}{λ} + 2) = \frac{c}{a} \frac{{(λ + μ)}^{2}}{λ μ}

λ μ b^{2} = ac {(λ + μ)}^{2} proved .

Commented by Rio Michael last updated on 02/Jul/19

c o o l i t h i n k t h a t s c o r r e c t .

Commented by Rio Michael last updated on 02/Jul/19

p l e a s e c h e c k m y s o l u t i o n .

α + β = - \frac{b}{a} a n d α β = \frac{c}{a}

- b = a (α + β)

a n d b^{2} = a^{2} {(α + β)}^{2} \dots \dots \dots . (i)

a n d α β = \frac{c}{a} \dots \dots \dots . (i i)

t a k i n g e q n (i) \times (i i)

\Rightarrow α β b^{2} = \frac{c}{a} \times a^{2} {(α + β)}^{2}

α β b^{2} = a c (α + β)

s i n c e \frac{α}{β} = \frac{λ}{μ} t h e n α = λ a n d β = μ

a n d s u b s t i t u t i n g w e g e t .

λ μ b^{2} = a c {(λ + μ)}^{2} p r o v e d .

p l e a s e m a k e m o r e c o r r e c t i o n s .

Commented by Prithwish sen last updated on 02/Jul/19

I think

\frac{α}{λ} = \frac{β}{μ} = k (let)

then α β b^{2} = ac {(α + β)}^{2} \Rightarrow k^{2} λ μ b^{2} = k^{2} ac {(λ + μ)}^{2}

Commented by Rio Michael last updated on 02/Jul/19

t h a t s a l s o a g o o d i d e a