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

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

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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 roots of $x^2 + qx + 1 = 0$,we have $\gamma + \delta = -q$ and $\gamma \delta = 1$.Consider the expression $(\alpha - \gamma)(\beta - \gamma)(\alpha + \delta)(\beta + \delta)$.Expanding the first part: $(\alpha - \gamma)(\beta - \gamma) = \alpha \beta - \gamma(\alpha + \beta) + \gamma^2 = 1 + p\gamma + \gamma^2$.Since $\gamma$ is a root of $x^2 + qx + 1 = 0$,we have $\gamma^2 + q\gamma + 1 = 0$,so $\gamma^2 + 1 = -q\gamma$.Thus,$1 + p\gamma + \gamma^2 = -q\gamma + p\gamma = \gamma(p - q)$.Similarly,expanding the second part: $(\alpha + \delta)(\beta + \delta) = \alpha \beta + \delta(\alpha + \beta) + \delta^2 = 1 - p\delta + \delta^2$.Since $\delta$ is a root of $x^2 + qx + 1 = 0$,we have $\delta^2 + q\delta + 1 = 0$,so $\delta^2 + 1 = -q\delta$.Thus,$1 - p\delta + \delta^2 = -q\delta - p\delta = -\delta(p + q)$.Multiplying these results: $\gamma(p - q) \cdot [-\delta(p + q)] = -\gamma \delta (p^2 - q^2) = -1 \cdot (p^2 - q^2) = q^2 - p^2$.

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 $q^2 - p^2$ =

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