If alpha, beta are the roots of x^2 + px + 1 | vedclass

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(A) Given that $\alpha, \beta$ are roots of $x^2 + px + 1 = 0$,we have $\alpha + \beta = -p$ and $\alpha \beta = 1$.Given that $\gamma, \delta$ are ro

Vedclass · Accepted Answer

$(\alpha - \gamma)(\beta - \gamma)(\alpha + \delta)(\beta + \delta)$

Vedclass · Answer

(A) Given that $\alpha, \beta$ are roots of $x^2 + px + 1 = 0$,we have $\alpha + \beta = -p$ and $\alpha \beta = 1$.Given that $\gamma, \delta$ are roots of $x^2 + qx + 1 = 0$,we have $\gamma + \delta = -q$ and $\gamma \delta = 1$.Consider the expression $(\alpha - \gamma)(\beta - \gamma)(\alpha + \delta)(\beta + \delta)$:$= (\alpha \beta - \gamma(\alpha + \beta) + \gamma^2)(\alpha \beta + \delta(\alpha + \beta) + \delta^2)$$= (1 + p\gamma + \gamma^2)(1 - p\delta + \delta^2)$.Since $\gamma$ is a root of $x^2 + qx + 1 = 0$,$\gamma^2 + q\gamma + 1 = 0$,so $1 + \gamma^2 = -q\gamma$.Similarly,$\delta^2 + 1 = -q\delta$.Substituting these into the expression:$= (-q\gamma + p\gamma)(-q\delta - p\delta)$$= \gamma(p - q) 	imes \delta(-p - q)$$= -\gamma \delta (p - q)(p + q)$$= -1(p^2 - q^2) = q^2 - p^2$.

If $\alpha, \beta$ are the roots of $x^2 + px + 1 = 0$ and $\gamma, \delta$ are the roots of $x^2 + qx + 1 = 0$,then $q^2 - p^2$ is equal to:

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