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

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(B) 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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$q^2 - p^2$

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(B) 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 $E = (\alpha - \gamma) (\beta - \gamma) (\alpha + \delta) (\beta + \delta)$. $E = [\alpha\beta - \gamma(\alpha + \beta) + \gamma^2] [\alpha\beta + \delta(\alpha + \beta) + \delta^2]$. Substituting the values: $E = [1 - \gamma(-p) + \gamma^2] [1 + \delta(-p) + \delta^2] = [1 + p\gamma + \gamma^2] [1 - p\delta + \delta^2]$. Since $\gamma^2 + q\gamma + 1 = 0$,we have $\gamma^2 + 1 = -q\gamma$. Since $\delta^2 + q\delta + 1 = 0$,we have $\delta^2 + 1 = -q\delta$. Substituting these: $E = [-q\gamma + p\gamma] [-q\delta - p\delta] = \gamma(p - q) \cdot \delta(-p - q) = -\gamma\delta(p - q)(p + q) = -1(p^2 - q^2) = q^2 - p^2$.

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

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