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(D) Given the quadratic equation $x^2 + 6x - 2 = 0$. Since $\alpha$ is a root,we have $\alpha^2 + 6\alpha - 2 = 0$,which implies $\alpha^2 = 2 - 6\alp

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All of these

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(D) Given the quadratic equation $x^2 + 6x - 2 = 0$. Since $\alpha$ is a root,we have $\alpha^2 + 6\alpha - 2 = 0$,which implies $\alpha^2 = 2 - 6\alpha$.For option $A$: $\alpha^2 + 5\alpha - 8 = (2 - 6\alpha) + 5\alpha - 8 = -\alpha - 6$. Since $\alpha + \beta = -6$,we have $\beta = -6 - \alpha$. Thus,option $A$ is correct.For option $C$: $\frac{2\alpha^2 + 12\alpha - 6}{\alpha} = \frac{2(\alpha^2 + 6\alpha) - 6}{\alpha} = \frac{2(2) - 6}{\alpha} = \frac{-2}{\alpha}$. Since $\alpha\beta = -2$,we have $\beta = \frac{-2}{\alpha}$. Thus,option $C$ is correct.For option $B$: From $\alpha\beta = -2$ and $\alpha + \beta = -6$,we have $\frac{\alpha + \beta}{\alpha\beta} = \frac{-6}{-2} = 3$,so $\frac{1}{\beta} + \frac{1}{\alpha} = 3$. This gives $\frac{\alpha + \beta}{\alpha\beta} = 3 \Rightarrow \frac{1}{\beta} = 3 - \frac{1}{\alpha} = \frac{3\alpha - 1}{\alpha}$,so $\beta = \frac{\alpha}{3\alpha - 1}$. Thus,option $B$ is correct.Since all options are equivalent to $\beta$,the correct answer is $D$.

If $\alpha$ is a root of the quadratic equation $x^2 + 6x - 2 = 0$,then another root $\beta$ is

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