If - - are the roots of x2-px-q-0 and - - are the roots of x2-px-r-0- then - - - - - is equal to

The correct option is B 1 α +β = p, nbsp;γ +δ = p αβ =q, nbsp;γδ = r α γβ γα δβ δ=α ^2 αδ γα +γδβ ^2 βδ γβ +γδ α ^2+pα r β ^ ...

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Standard XII

Physics

2nd Law of Thermodynamics

If α, β are...

Question

If $α, β$ are the roots of $x^{2} + p x + q = 0$ and $γ, δ$ are the roots of $x^{2} + p x - r = 0$ , then $\frac{α - γ}{β - γ} \cdot \frac{α - δ}{β - δ}$ is equal to

1

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- 1

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\frac{q + r}{p + r}

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\frac{q - r}{p - r}

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Solution

The correct option is B $1$
$α + β = - p, γ + δ = - p$
$α β = q, γ δ = - r$
$\frac{α - γ}{β - γ} \frac{α - δ}{β - δ} = \frac{α^{2} - α δ - γ α + γ δ}{β^{2} - β δ - γ β + γ δ}$
$\frac{α^{2} + p α - r}{β^{2} + p β - r} = \frac{- q - r}{- q - r} = 1$

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Similar questions

Q. If

α, β

are roots of the equation

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

and

γ, δ

are roots of

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

then the value of

(α - γ) (α - δ)

is-

Q. If

α, β

be the roots

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

and

γ, δ

be the roots of

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

, then

\frac{(α - γ) (α - δ)}{(β - γ) (β - δ)} =

Q. If

α

β

are the roots of quadratic equation

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

and

γ

δ

are the roots of

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

then

(α - γ)

(α - δ)

is equal to:

Q. lf

α, β

are the roots of

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

and

γ, δ

that of

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

, then

(α - γ) (β - γ) (α - δ) (β - δ) =

Q. If

α

and

β

be the roots of

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

and

γ, δ

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If α,β are the roots of x2+px+q=0 and γ,δ are the roots of x2+px−r=0, then α−γβ−γ⋅α−δβ−δ is equal to

The correct option is B 1α+β=−p,γ+δ=−pαβ=q,γδ=−rα−γβ−γα−δβ−δ=α2−αδ−γα+γδβ2−βδ−γβ+γδα2+pα−rβ2+pβ−r=−q−r−q−r=1

If $α, β$ are the roots of $x^{2} + p x + q = 0$ and $γ, δ$ are the roots of $x^{2} + p x - r = 0$ , then $\frac{α - γ}{β - γ} \cdot \frac{α - δ}{β - δ}$ is equal to

The correct option is B $1$
$α + β = - p, γ + δ = - p$
$α β = q, γ δ = - r$
$\frac{α - γ}{β - γ} \frac{α - δ}{β - δ} = \frac{α^{2} - α δ - γ α + γ δ}{β^{2} - β δ - γ β + γ δ}$
$\frac{α^{2} + p α - r}{β^{2} + p β - r} = \frac{- q - r}{- q - r} = 1$