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 $x^2 + x + 1 = 0$.From the properties of roots,$\alpha + \beta = -1$ and $\alpha \beta = 1$.Now,$

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$1$

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(C) Given that $\alpha$ and $\beta$ are the roots of $x^2 + x + 1 = 0$.From the properties of roots,$\alpha + \beta = -1$ and $\alpha \beta = 1$.Now,$\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ are the roots of $x^2 + px + q = 0$.The sum of the roots is given by $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -p$.Simplifying the sum: $\frac{\alpha^2 + \beta^2}{\alpha \beta} = -p$.Using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$,we get:$\frac{(\alpha + \beta)^2 - 2\alpha \beta}{\alpha \beta} = -p$.Substituting the values $\alpha + \beta = -1$ and $\alpha \beta = 1$:$\frac{(-1)^2 - 2(1)}{1} = -p$.$\frac{1 - 2}{1} = -p$.$-1 = -p$,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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