If α and β are the roots of the ax^2 + 2bx + c = 0 and α + δ and β + δ are the roots of Ax^2 + 2Bx + C = 0 for some constant δ then prove that ((b^2 − ac)/a^2 ) = ((B^2 − AC)/A^2 ) .


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Question Number 202019 by MATHEMATICSAM last updated on 18/Dec/23

$If α and β are the roots of the$ ${a x}^{2} + 2 b x + c = 0 and α + δ and β + δ are$ $the roots of {A x}^{2} + 2 B x + C = 0 for some$ $constant δ then prove that$ $\frac{b^{2} - a c}{a^{2}} = \frac{B^{2} - A C}{A^{2}} .$
	Answered by esmaeil last updated on 18/Dec/23

	$α - β = \frac{\sqrt{b^{2} - a c}}{∣ a ∣} = (α + δ) - (β + δ) =$ $\frac{\sqrt{B^{2} - A C}}{∣ A ∣} \to \frac{b^{2} - a c}{a^{2}} = \frac{B^{2} - A C}{A^{2}}$ $α - β = \frac{\sqrt{δ^{^{'}}}}{∣ a ∣}$
	Commented by MM42 last updated on 18/Dec/23

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	Answered by Rasheed.Sindhi last updated on 19/Dec/23

	$A n o t h e r w a y$ $∙ α & β a r e t h e r o o t s o f {a x}^{2} + 2 b x + c = 0$ $α + β = - \frac{2 b}{a}, α β = \frac{c}{a}$ $∙ I f t h e r o o t s a r e α + δ & β + δ (δ i s f i x e d c o n s t a n t)$ $T h e e q u a t i o n w i l l b e :$ $(α + δ) + (β + δ) = α + β + 2 δ = - \frac{2 b}{a} + 2 δ = - \frac{2 B}{A}$ $\Rightarrow \frac{B}{A} = \frac{- 2 b + 2 a δ}{- 2 a}$ $(α + δ) . (β + δ) = α β + δ^{2} + (α + β) δ = \frac{c}{a} - \frac{2 b δ}{a} + δ^{2} = \frac{C}{A}$ $\Rightarrow \frac{C}{A} = \frac{c - 2 b δ + a δ^{2}}{a}$ $\frac{B^{2} - A C}{A^{2}} = {(\frac{B}{A})}^{2} - \frac{C}{A} = {(\frac{- 2 b + 2 a δ}{- 2 a})}^{2} - \frac{c - 2 b δ + a δ^{2}}{a}$ $= \frac{4 b^{2} + 4 a^{2} δ^{2} - 8 a b δ}{4 a^{2}} - \frac{c - 2 b δ + a δ^{2}}{a}$ $= \frac{4 b^{2} + 4 a^{2} δ^{2} - 8 a b δ - 4 a c + 8 a b δ - 4 a^{2} δ^{2}}{4 a^{2}}$ $= \frac{4 b^{2} - 4 a c}{4 a^{2}} = \frac{b^{2} - a c}{a^{2}}$ $Q E D$
	Answered by Rasheed.Sindhi last updated on 19/Dec/23

	$\frac{b^{2} - a c}{a^{2}} = \frac{B^{2} - A C}{A^{2}} .$ ${(\frac{b}{a})}^{2} - \frac{c}{a} = {(\frac{B}{A})}^{2} - \frac{C}{A}$ $\frac{1}{4} {(\frac{- 2 b}{a})}^{2} - \frac{c}{a} = \frac{1}{4} {(\frac{- 2 B}{A})}^{2} - \frac{C}{A}$ $\frac{1}{4} {(α + β)}^{2} - α β = \frac{1}{4} {((α + δ) + (β + δ))}^{2} - (α + δ) (β + δ)$ $\frac{1}{4} {(α + β)}^{2} - α β = \frac{1}{4} {(α + β + 2 δ)}^{2} - (α β + (α + β) δ + δ^{2})$ ${(α + β)}^{2} - 4 α β = {(α + β + 2 δ)}^{2} - 4 (α β + (α + β) δ + δ^{2})$ $= {(α + β)}^{2} + 4 (α + β) δ + 4 δ^{2} - 4 α β - 4 (α + β) δ - 4 δ^{2}$ $= {(α + β)}^{2} - 4 α β$ $l h s = r h s$ $p r o v e d$
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