If - are the roots of x2-2px-q-0 and - are roots f x2-2rx-s-0 and - are in A-P- then

The correct option is B s q=r^2 p^2 We have, #160;α +β =2p;αβ =q,γ +δ =2r and #160;γδ =s . #160;∵α ,β ,γ ,δ are in A.P. #160;∴β ...

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Standard X

Mathematics

Relationship between Zeroes and Coefficients of a Polynomial

If α,β are ...

Question

If $α, β$ are the roots of $x^{2} - 2 p x + q = 0$ and $γ, δ$ are roots f $x^{2} - 2 r x + s = 0$ and $α, β, γ, δ$ are in $A . P .,$ then

p - q = r^{2} - s^{2}

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s - q = r^{2} - p^{2}

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r - s = p^{2} - q^{2}

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None of these

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Solution

The correct option is B $s - q = r^{2} - p^{2}$
We have, $α + β = 2 p; α β = q, γ + δ = 2 r$ and $γ δ = s$ .
$∵ α, β, γ, δ$ are in $A . P .$
$∴ β - α = δ - γ \Rightarrow {(β - α)}^{2} = {(δ - γ)}^{2}$
$\Rightarrow {(β - α)}^{2} - 4 β α = {(δ + γ)}^{2} - 4 δ γ$
$\Rightarrow 4 p^{2} - 4 q = 4 r^{2} - 4 s;$ or $s - q = r^{2} - p^{2} .$

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If α,β are the roots of x2−2px+q=0 and γ,δ are roots f x2−2rx+s=0 and α,β,γ,δ are in A.P., then

The correct option is B s−q=r2−p2We have, α+β=2p;αβ=q,γ+δ=2r and γδ=s.∵α,β,γ,δ are in A.P.∴β−α=δ−γ⇒(β−α)2=(δ−γ)2⇒(β−α)2−4βα=(δ+γ)2−4δγ⇒4p2−4q=4r2−4s; or s−q=r2−p2.

If $α, β$ are the roots of $x^{2} - 2 p x + q = 0$ and $γ, δ$ are roots f $x^{2} - 2 r x + s = 0$ and $α, β, γ, δ$ are in $A . P .,$ then