If the roots of the equation x^2 + x + 1 = 0 | vedclass

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(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 + x + 1 = 0$,we have $\alpha + \beta = -1$ and $\alpha \beta = 1$.The roots of the equation

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$1$

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(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 + x + 1 = 0$,we have $\alpha + \beta = -1$ and $\alpha \beta = 1$.The roots of the equation $x^2 + px + q = 0$ are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.By the relation between roots and coefficients,the sum of the roots is $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -p$.Thus,$-p = \frac{\alpha^2 + \beta^2}{\alpha \beta}$.Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$,we get:$-p = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta} = \frac{(-1)^2 - 2(1)}{1} = \frac{1 - 2}{1} = -1$.Therefore,$-p = -1$,which implies $p = 1$.

If the roots of the equation $x^2 + x + 1 = 0$ are $\alpha$ and $\beta$,and the roots of the equation $x^2 + px + q = 0$ are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$,then $p$ is equal to:

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