If alpha and beta are the roots of the equati | vedclass

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(B) Given $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Expanding the expression $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2)$:$= 1

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Positive

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(B) Given $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Expanding the expression $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2)$:$= 1 + (\alpha + \beta) + (\alpha^2 + \beta^2) + \alpha\beta + \alpha\beta(\alpha + \beta) + (\alpha\beta)^2$$= 1 + (\alpha + \beta) + ((\alpha + \beta)^2 - 2\alpha\beta) + \alpha\beta + \alpha\beta(\alpha + \beta) + (\alpha\beta)^2$$= 1 + (\alpha + \beta) + (\alpha + \beta)^2 - \alpha\beta + \alpha\beta(\alpha + \beta) + (\alpha\beta)^2$Substituting the values:$= 1 - \frac{b}{a} + \frac{b^2}{a^2} - \frac{c}{a} - \frac{bc}{a^2} + \frac{c^2}{a^2}$$= \frac{a^2 - ab + b^2 - ac - bc + c^2}{a^2}$$= \frac{(a-b)^2 + (b-c)^2 + (c-a)^2}{2a^2}$Since $a, b, c$ are distinct,the numerator is always positive and $2a^2 > 0$,so the expression is always positive.

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$ ($a \ne 0$; $a, b, c$ being distinct),then $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2) = $

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