If - - are the roots of a x2-2 b x-c-0 and a- - are the roots of A x2-2 B x-C-0 then b2-a c-B2-A c is

The correct option is B ( a/A)2 (α β)^2 = (α+β)^2 4αβ =[(α+δ) (β+δ)]^2 = (α+δ+β+δ)^2 4(α+δ)(β+δ) = gt; b^2 – ac/B^2 Ac = a^2/A^2 ...

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Byju's Answer

Standard XI

Mathematics

Equations Which Can Be Solved by Substituting t = tan(X/2)

If α, β are t...

Question

If α, β are the roots of a $x^{2}$ +2bx+c=0 and a+δ, β+δ are the roots of A $x^{2}$ + 2Bx + C=0 then $\frac{b^{2} - a c}{B^{2} - A c}$ is

a/A

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( a/A)²

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A/a

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( A/a)²

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Solution

The correct option is B
( a/A)²

${(α - β)}^{2}$ = ${(α + β)}^{2}$ - 4αβ = ${[(α + δ) - (β + δ)]}^{2}$ = ${(α + δ + β + δ)}^{2}$ -4(α+δ)(β+δ)

=> $\frac{b^{2} - a c}{B^{2} - A c}$ = $\frac{a^{2}}{A^{2}}$

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Similar questions

If $α, β$ are the roots of the equation a $x^{2}$ +2bx +c = 0 and $α + δ$ , $β + δ$ are the roots of A $x^{2}$ + 2Bx + C= 0 , then $\frac{b^{2} - a c}{B^{2} - A C}$ =

Q. If

α, β

be the roots of

a x^{2} + 2 b x + c = 0

and

α + δ, β + δ

be those of

A x^{2} + 2 B x + C = 0

for some constant

δ

, then prove that

\frac{b^{2} - a c}{B^{2} - A C} = {(\frac{a}{A})}^{2}

Q. If

α, β

are the roots of

a x^{2} + 2 b x + c = 0

and

α + δ

β δ

are the roots of

A x^{2} + 2 B x + C = 0

, then

Q. If

α, β

are the roots of

a x^{2} + b x + c = 0

and

α + β, α^{2} + β^{2}, α^{3} + β^{3}

are in G.P., where

△ = b^{2} - 4 a c

, then

Q. If

α, β, γ, δ

are in GP where

α, β

are roots of the equation

a x^{2} + 2 b x + c = 0

and

γ, δ

are roots of the equation

p x^{2} + 2 q x + r = 0,

then

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If α, β are the roots of ax2 +2bx+c=0 and a+δ, β+δ are the roots of Ax2 + 2Bx + C=0 then b2–acB2−Ac is

The correct option is B ( a/A)2 (α−β)2 = (α+β)2 - 4αβ =[(α+δ)−(β+δ)]2 = (α+δ+β+δ)2-4(α+δ)(β+δ) => b2–acB2−Ac = a2A2

If α, β are the roots of a $x^{2}$ +2bx+c=0 and a+δ, β+δ are the roots of A $x^{2}$ + 2Bx + C=0 then $\frac{b^{2} - a c}{B^{2} - A c}$ is

The correct option is B
( a/A)²

${(α - β)}^{2}$ = ${(α + β)}^{2}$ - 4αβ = ${[(α + δ) - (β + δ)]}^{2}$ = ${(α + δ + β + δ)}^{2}$ -4(α+δ)(β+δ)

=> $\frac{b^{2} - a c}{B^{2} - A c}$ = $\frac{a^{2}}{A^{2}}$