If - are the roots of x-2-px-q-0 and - - are the roots of x-2-px-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 ...

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Standard XIII

Mathematics

Relations between Roots and Coefficients

If α,β are th...

Question

If $\alpha,\beta$ are the roots of $x^2+px-q=0$ and $\gamma, \delta$ are the roots of $x^2+px+r=0$, then the value of $\dfrac{(\alpha-\gamma)(\alpha-\delta)}{(\beta-\gamma)(\beta-\delta)}$ is

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Solution

$\alpha,\beta$ are the roots of $x^2+px-q=0$
Sum of the roots,
$\alpha+\beta=-p$

$\gamma, \delta$ are the roots of $x^2+px+r=0$
Sum of the roots,
$\gamma+\delta=-p$

Now,
$\alpha+\beta = \gamma+\delta$
$\Rightarrow \alpha-\gamma=\delta - \beta \cdots(1)\\
\Rightarrow \alpha - \delta = \gamma - \beta \cdots(2)$

Using equation $(1)$ and $(2)$,
$\dfrac{(\alpha-\gamma)(\alpha-\delta)}{(\beta-\gamma)(\beta-\delta)} = \dfrac{(\delta - \beta)(\gamma - \beta)}{(\beta-\gamma)(\beta-\delta)} = 1$

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If \(\alpha,\beta\) are the roots of \(x^2+px-q=0\) and \(\gamma, \delta\) are the roots of \(x^2+px+r=0\), then the value of \(\dfrac{(\alpha-\gamma)(\alpha-\delta)}{(\beta-\gamma)(\beta-\delta)}\) is