The problem states that $\alpha$ and $\beta$ are the roots of the quadratic equation $3x^2 - 6x - 9 = 0$. We need to find the quadratic equation whose roots are $\frac{1}{\alpha - 1}$ and $\frac{1}{\beta - 1}$.

1. Problem Description

The problem states that $\alpha$ and $\beta$ are the roots of the quadratic equation $3x^2 - 6x - 9 = 0$. We need to find the quadratic equation whose roots are $\frac{1}{\alpha - 1}$ and $\frac{1}{\beta - 1}$.

2. Solution Steps

First, let's simplify the given quadratic equation: $3x^2 - 6x - 9 = 0$. We can divide the equation by 3 to get $x^2 - 2x - 3 = 0$.

Since $\alpha$ and $\beta$ are the roots of this equation, we have:

Sum of the roots: $\alpha + \beta = -\frac{-2}{1} = 2$

Product of the roots: $\alpha \beta = \frac{-3}{1} = -3$

Now, let's find the sum and product of the new roots $\frac{1}{\alpha - 1}$ and $\frac{1}{\beta - 1}$.

Sum of the new roots:

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

Substituting the values of $\alpha + \beta$ and $\alpha\beta$, we get:

$S = \frac{2 - 2}{-3 - 2 + 1} = \frac{0}{-4} = 0$

Product of the new roots:

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

Substituting the values of $\alpha + \beta$ and $\alpha\beta$, we get:

$P = \frac{1}{-3 - 2 + 1} = \frac{1}{-4} = -\frac{1}{4}$

The general form of a quadratic equation with roots $r_1$ and $r_2$ is given by:

$x^2 - (r_1 + r_2)x + r_1 r_2 = 0$

In our case, the new roots are $\frac{1}{\alpha - 1}$ and $\frac{1}{\beta - 1}$, so the quadratic equation is:

$x^2 - Sx + P = 0$

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

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

Multiplying the equation by 4 to eliminate the fraction, we get:

$4x^2 - 1 = 0$

3. Final Answer

The quadratic equation whose roots are $\frac{1}{\alpha - 1}$ and $\frac{1}{\beta - 1}$ is $4x^2 - 1 = 0$.