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

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Question

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

Vedclass · Accepted Answer

Positive

Vedclass · Answer

(B) Given $\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$.Consider the expression $E = (1 + \alpha + \alpha^2)(1 + \beta + \beta^2)$.Expanding the product:$E = 1 + (\alpha + \beta) + (\alpha^2 + \beta^2) + \alpha \beta + \alpha \beta(\alpha + \beta) + (\alpha \beta)^2$Using $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$:$E = 1 + (\alpha + \beta) + ((\alpha + \beta)^2 - 2\alpha \beta) + \alpha \beta + \alpha \beta(\alpha + \beta) + (\alpha \beta)^2$Substituting the values of $\alpha + \beta$ and $\alpha \beta$:$E = 1 - \frac{b}{a} + (\frac{b^2}{a^2} - 2\frac{c}{a}) + \frac{c}{a} + \frac{c}{a}(-\frac{b}{a}) + \frac{c^2}{a^2}$$E = \frac{a^2 - ab + b^2 - ac - bc + c^2}{a^2} = \frac{a^2 + b^2 + c^2 - ab - bc - ca}{a^2}$$E = \frac{(a-b)^2 + (b-c)^2 + (c-a)^2}{2a^2}$Since the sum of squares is non-negative and $a 
eq 0$,the expression is positive.

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$ ($a \neq 0$; $a, b, c \in \mathbb{R}$),then $(1 + \alpha + \alpha^2)(1 + \beta + \beta^2)$ is . . . .

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