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(C) Given that $\alpha$ and $\beta$ are the roots of the equation $x^2 - 3x + 1 = 0$.From the relation between roots and coefficients,we have $\alpha

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$x^2 - x - 1 = 0$

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(C) Given that $\alpha$ and $\beta$ are the roots of the equation $x^2 - 3x + 1 = 0$.From the relation between roots and coefficients,we have $\alpha + \beta = 3$ and $\alpha \beta = 1$.Let the new roots be $x_1 = \frac{1}{\alpha - 2}$ and $x_2 = \frac{1}{\beta - 2}$.The sum of the new roots $S = x_1 + x_2 = \frac{1}{\alpha - 2} + \frac{1}{\beta - 2} = \frac{\alpha + \beta - 4}{\alpha \beta - 2(\alpha + \beta) + 4}$.Substituting the values,$S = \frac{3 - 4}{1 - 2(3) + 4} = \frac{-1}{1 - 6 + 4} = \frac{-1}{-1} = 1$.The product of the new roots $P = x_1 x_2 = \frac{1}{(\alpha - 2)(\beta - 2)} = \frac{1}{\alpha \beta - 2(\alpha + \beta) + 4}$.Substituting the values,$P = \frac{1}{1 - 2(3) + 4} = \frac{1}{1 - 6 + 4} = \frac{1}{-1} = -1$.The required quadratic equation is $x^2 - Sx + P = 0$,which gives $x^2 - (1)x + (-1) = 0$,or $x^2 - x - 1 = 0$.

If $\alpha$ and $\beta$ are roots of $x^2 - 3x + 1 = 0$,then the equation whose roots are $\frac{1}{\alpha - 2}$ and $\frac{1}{\beta - 2}$ is

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