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Relations between Roots and Coefficients : Higher Order Equations

lf α, β are...

Question

lf $α, β$ are the roots of $x^{2} - p x + q = 0$ , then the equation whose roots are $α β + α + β, α β - α - β$ , is:

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

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x^{2} - 2 q x + p^{2} + q^{2} = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} - 2 q x + q^{2} - p^{2} = 0

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x^{2} + 2 q x + q^{2} - p^{2} = 0

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Solution

The correct option is C $x^{2} - 2 q x + q^{2} - p^{2} = 0$
As $α, β$ are roots of $x^{2} - p x + q = 0,$ then
$S_{1} = α + β = p$
$S_{2} = α β = q$
$∴ α β + α + β + α β - α - β$ $= 2 α β = 2 q$
$(α β + α + β) (α β - α - β) = (α^{2} β^{2} - {(α + β)}^{2}) = (q^{2} - p^{2})$
$∴$ Required equation is $x^{2} - (2 q) x + (q^{2} - p^{2}) = 0$

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Similar questions

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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 $(α - γ) (α + δ) (β - γ) (β + δ)$

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

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 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.

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lf α, β are the roots of x2−px+q=0, then the equation whose roots are αβ+α+β, αβ−α−β, is:

The correct option is C x2−2qx+q2−p2=0As α,β are roots of x2−px+q=0, thenS1=α+β=pS2=αβ=q∴αβ+α+β+αβ−α−β=2αβ=2q(αβ+α+β)(αβ−α−β)=(α2β2−(α+β)2)=(q2−p2)∴ Required equation is x2−(2q)x+(q2−p2)=0

lf $α, β$ are the roots of $x^{2} - p x + q = 0$ , then the equation whose roots are $α β + α + β, α β - α - β$ , is: