Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 + 2x - 4 = 0$, we need to find the quadratic equation whose roots are $\frac{3}{\beta}$ and $\frac{3}{\alpha}$.

1. Problem Description

Given that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2 + 2x - 4 = 0$, we need to find the quadratic equation whose roots are $\frac{3}{\beta}$ and $\frac{3}{\alpha}$.

2. Solution Steps

First, we find the sum and product of the roots of the given equation $x^2 + 2x - 4 = 0$.

The sum of the roots is $\alpha + \beta = -\frac{2}{1} = -2$.

The product of the roots is $\alpha \beta = \frac{-4}{1} = -4$.

Now, we want to find the equation whose roots are $\frac{3}{\beta}$ and $\frac{3}{\alpha}$.

The sum of the new roots is $\frac{3}{\beta} + \frac{3}{\alpha} = 3(\frac{1}{\beta} + \frac{1}{\alpha}) = 3(\frac{\alpha + \beta}{\alpha \beta}) = 3(\frac{-2}{-4}) = 3(\frac{1}{2}) = \frac{3}{2}$.

The product of the new roots is $\frac{3}{\beta} \cdot \frac{3}{\alpha} = \frac{9}{\alpha \beta} = \frac{9}{-4} = -\frac{9}{4}$.

A quadratic equation with roots $r_1$ and $r_2$ can be written as $x^2 - (r_1 + r_2)x + r_1 r_2 = 0$.

Therefore, the equation with roots $\frac{3}{\beta}$ and $\frac{3}{\alpha}$ is given by

$x^2 - (\frac{3}{2})x + (-\frac{9}{4}) = 0$

$x^2 - \frac{3}{2}x - \frac{9}{4} = 0$

Multiplying by 4 to eliminate fractions, we get

$4x^2 - 6x - 9 = 0$

3. Final Answer

$4x^2 - 6x - 9 = 0$