If alpha and beta are the roots of the equati | vedclass

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(A) Let $y = \frac{1}{x - 2}$. Then $x - 2 = \frac{1}{y}$,so $x = \frac{1}{y} + 2 = \frac{1 + 2y}{y}$.Substitute $x = \frac{2y + 1}{y}$ into the origi

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$x^2 - x - 1 = 0$

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(A) Let $y = \frac{1}{x - 2}$. Then $x - 2 = \frac{1}{y}$,so $x = \frac{1}{y} + 2 = \frac{1 + 2y}{y}$.Substitute $x = \frac{2y + 1}{y}$ into the original equation $x^2 - 3x + 1 = 0$:$(\frac{2y + 1}{y})^2 - 3(\frac{2y + 1}{y}) + 1 = 0$$\frac{4y^2 + 4y + 1}{y^2} - \frac{6y + 3}{y} + 1 = 0$Multiply by $y^2$:$(4y^2 + 4y + 1) - y(6y + 3) + y^2 = 0$$4y^2 + 4y + 1 - 6y^2 - 3y + y^2 = 0$$-y^2 + y + 1 = 0$$y^2 - y - 1 = 0$.Thus,the required equation is $x^2 - x - 1 = 0$.

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

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