If - - are the roots of x2-p x-1-0- - - the roots of x2-q x-1-0- then -A- p 2B- q2 C- p2-q2D- q2-p2

The correct option is D q2 p2 α+β= p , αβ=1 also γ^2+1= qγ and δ^2+1= qδ (∝ γ)( β γ)(∝+ δ)( β +δ)= (αβ (α+β)γ+γ^2)(αβ+(α+β)δ+δ^2) =(γ^2+pγ+1)(δ^2 pδ+1)= ...

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Byju's Answer

Standard XI

Mathematics

Properties of Inequalities

If ∝, β are t...

Question

If ∝, β are the roots of $x^{2}$ + px + 1 = 0, γ, δ the roots of $x^{2}$ + qx + 1 = 0, then (∝ - γ)( β -γ)(∝+ δ)( β +δ) =

p²

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q²

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p² - q²

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q² - p²

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Solution

The correct option is D
q² - p²

α+β=-p , αβ=1 also $γ^{2}$ +1=-qγ and $δ^{2}$ +1= -qδ

(∝ - γ)( β -γ)(∝+ δ)( β +δ)= (αβ-(α+β)γ+ $γ^{2}$ )(αβ+(α+β)δ+ $δ^{2}$ )

=( $γ^{2}$ +pγ+1)( $δ^{2}$ -pδ+1)= (pγ-qγ)(-pδ-qδ)=( $q^{2}$ -p2)λδ=( $q^{2}$ - $p^{2}$ )

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Similar questions

Q. If α, β are the roots of the equation

x^{2} + p x + 1 = 0; γ, δ

the roots of the equation

x^{2} + q x + 1 = 0, t h e n (α - γ) (α + δ) (β - γ) (β + δ) =

(a)

q^{2} - p^{2}

(b)

p^{2} - q^{2}

(c)

p^{2} + q^{2}

(d) none of these.

Q. If

α, β

roots of

x^{2} + p x + 1 = 0

and

γ, δ

are the roots of

x^{2} + q x + 1 = 0

then prove that

(α - γ) (β - γ) (α + δ) (β + δ) = q^{2} - p^{2}

Q. If

α, β

be the roots of

x^{2} + p x + 1 = 0

and

γ, δ

be the roots of

x^{2} + q x + 1 = 0

, then

(α - γ) (β - γ) (α + δ) (β + δ) =

Q. If

α

and

β

are roots of

x^{2} + p x + 1 = 0

, and

γ

and

δ

are the roots of

x^{2} + q x + 1 = 0

show that

q^{2} - p^{2} - (α - γ) (β - γ) (α + δ) (β + δ) = 0

If $α, β$ are the roots of the equation $x^{2} + p x + 1 = 0; γ, δ$ the roots of the equation $x^{2} + p x + 1 = 0;$ then $(α - γ) (α + δ) (β - γ) (β + δ)$

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If ∝, β are the roots of x2 + px + 1 = 0, γ, δ the roots of x2 + qx + 1 = 0, then (∝ - γ)( β -γ)(∝+ δ)( β +δ) =

The correct option is D q2 - p2 α+β=-p , αβ=1 also γ2+1=-qγ and δ2+1= -qδ (∝ - γ)( β -γ)(∝+ δ)( β +δ)= (αβ-(α+β)γ+γ2)(αβ+(α+β)δ+δ2) =(γ2+pγ+1)(δ2-pδ+1)= (pγ-qγ)(-pδ-qδ)=(q2-p2)λδ=(q2-p2)

If ∝, β are the roots of $x^{2}$ + px + 1 = 0, γ, δ the roots of $x^{2}$ + qx + 1 = 0, then (∝ - γ)( β -γ)(∝+ δ)( β +δ) =

The correct option is D
q² - p²

α+β=-p , αβ=1 also $γ^{2}$ +1=-qγ and $δ^{2}$ +1= -qδ

(∝ - γ)( β -γ)(∝+ δ)( β +δ)= (αβ-(α+β)γ+ $γ^{2}$ )(αβ+(α+β)δ+ $δ^{2}$ )

=( $γ^{2}$ +pγ+1)( $δ^{2}$ -pδ+1)= (pγ-qγ)(-pδ-qδ)=( $q^{2}$ -p2)λδ=( $q^{2}$ - $p^{2}$ )