If-and-are-the-roots-of-the-ax-2-2bx-c-0-and-and-are-the-roots-of-Ax-2-2Bx-C-0-for-some-constant-then-prove-that-b-2-ac-a-2-B-2-AC-A-2-

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

⋚

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