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(D) Given the quadratic equation $x^2 + 6x - 2 = 0$ with roots $\alpha$ and $\beta$.From the properties of roots,$\alpha + \beta = -6$ and $\alpha\bet

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(D) Given the quadratic equation $x^2 + 6x - 2 = 0$ with roots $\alpha$ and $\beta$.From the properties of roots,$\alpha + \beta = -6$ and $\alpha\beta = -2$.$1$. Since $\alpha$ is a root,$\alpha^2 + 6\alpha - 2 = 0$. Thus,$\alpha^2 = 2 - 6\alpha$.Substituting this into option $(A)$: $\alpha^2 + 5\alpha - 8 = (2 - 6\alpha) + 5\alpha - 8 = -\alpha - 6$. Since $\beta = -6 - \alpha$,option $(A)$ is correct.$2$. From $\alpha\beta = -2$,we have $\beta = -2/\alpha$. Substituting $\alpha^2 + 6\alpha - 2 = 0$ into the numerator of option $(C)$: $\frac{2(\alpha^2 + 6\alpha - 2) - 2}{\alpha} = \frac{2(0) - 2}{\alpha} = -2/\alpha$. Thus,option $(C)$ is correct.$3$. From $\alpha + \beta = -6$ and $\alpha\beta = -2$,we have $\frac{\alpha + \beta}{\alpha\beta} = \frac{-6}{-2} = 3$. This implies $\frac{1}{\beta} + \frac{1}{\alpha} = 3$,so $\frac{\alpha + \beta}{\alpha\beta} = 3$. Rearranging for $\beta$: $\frac{1}{\beta} = 3 - \frac{1}{\alpha} = \frac{3\alpha - 1}{\alpha}$,which gives $\beta = \frac{\alpha}{3\alpha - 1}$. 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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