The problem asks us to find a quadratic equation whose roots are $\frac{1}{\alpha-1}$ and $\frac{1}{\beta-1}$, where $\alpha$ and $\beta$ are the roots of the given quadratic equation $3x^2 - 6x - 9 = 0$.

1. Problem Description

The problem asks us to find a quadratic equation whose roots are $\frac{1}{\alpha-1}$ and $\frac{1}{\beta-1}$, where $\alpha$ and $\beta$ are the roots of the given quadratic equation $3x^2 - 6x - 9 = 0$.

2. Solution Steps

First, we simplify the given quadratic equation:

$3x^2 - 6x - 9 = 0$

$x^2 - 2x - 3 = 0$

From Vieta's formulas, we know that for a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-\frac{b}{a}$ and the product of the roots is $\frac{c}{a}$. Thus, for the equation $x^2 - 2x - 3 = 0$, we have:

$\alpha + \beta = -\frac{-2}{1} = 2$

$\alpha \beta = \frac{-3}{1} = -3$

We want to find a quadratic equation with roots $\frac{1}{\alpha-1}$ and $\frac{1}{\beta-1}$. Let $S$ be the sum of the new roots and $P$ be the product of the new roots.

$S = \frac{1}{\alpha-1} + \frac{1}{\beta-1} = \frac{(\beta-1) + (\alpha-1)}{(\alpha-1)(\beta-1)} = \frac{\alpha + \beta - 2}{\alpha \beta - (\alpha + \beta) + 1} = \frac{2 - 2}{-3 - 2 + 1} = \frac{0}{-4} = 0$

$P = \frac{1}{\alpha-1} \cdot \frac{1}{\beta-1} = \frac{1}{(\alpha-1)(\beta-1)} = \frac{1}{\alpha \beta - (\alpha + \beta) + 1} = \frac{1}{-3 - 2 + 1} = \frac{1}{-4} = -\frac{1}{4}$

A quadratic equation with roots $\frac{1}{\alpha-1}$ and $\frac{1}{\beta-1}$ can be written as $x^2 - Sx + P = 0$. Substituting the values of $S$ and $P$, we get:

$x^2 - 0x - \frac{1}{4} = 0$

$x^2 - \frac{1}{4} = 0$

$4x^2 - 1 = 0$

3. Final Answer

The required quadratic equation is $4x^2 - 1 = 0$.