If - - are the roots of x2-p x-q-0 and - - are the roots of x2-p x-r-0- then the value of - is

Α,β are the roots of x^2+px q=0 Sum of the roots, α+β= p nbsp; γ, δ are the roots of x^2+px+r=0 Sum of the roots, γ+δ= p Now, α+β = γ+δ ⇒α γ=δ β⋯(1) ⇒α δ ...

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Standard XII

Mathematics

Higher Order Equations

If α, β are t...

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 the value of $\frac{(α - γ) (α - δ)}{(β - γ) (β - δ)}$ is

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Solution

$α, β$ are the roots of $x^{2} + p x - q = 0$
Sum of the roots,
$α + β = - p$
$γ, δ$ are the roots of $x^{2} + p x + r = 0$
Sum of the roots,
$γ + δ = - p$
Now,
$α + β = γ + δ$
$\Rightarrow α - γ = δ - β \dots (1) \Rightarrow α - δ = γ - β \dots (2)$

Using equation $(1)$ and $(2)$ ,
$\frac{(α - γ) (α - δ)}{(β - γ) (β - δ)} = \frac{(δ - β) (γ - β)}{(β - γ) (β - δ)} = 1$

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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 the value of (α−γ)(α−δ)(β−γ)(β−δ) is

α,β are the roots of x2+px−q=0 Sum of the roots, α+β=−p γ,δ are the roots of x2+px+r=0 Sum of the roots, γ+δ=−p Now, α+β=γ+δ ⇒α−γ=δ−β⋯(1)⇒α−δ=γ−β⋯(2) Using equation (1) and (2), (α−γ)(α−δ)(β−γ)(β−δ)=(δ−β)(γ−β)(β−γ)(β−δ)=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 the value of $\frac{(α - γ) (α - δ)}{(β - γ) (β - δ)}$ is