Find the set S that denotes the set of all values of "$$\alpha$$" for which the roots of the equation $$(1 - \alpha)x^2 - 6 \alpha x + 8 \alpha = 0$$ is greater than 2.

$$f\left(x\right)=\left(1-\alpha\ \right)x^2-6\alpha x+8\alpha\ =0$$

Now roots are greater than 2 therefore,

$$-\frac{b}{2a}>2$$

$$f\left(2\right)>0$$

D>0

$$-\frac{b}{2a}>0$$

$$\frac{6\alpha}{2\left(1-\alpha\ \right)}>0$$

$$\frac{\alpha}{\alpha\ -1}<0$$

$$\alpha\ \in\ \left(0,1\right)$$

f(2)>0

$$\left(1-\alpha\ \right)4-12\alpha\ +8\alpha\ >0$$

$$4-8\alpha\ >0$$

$$\alpha\ <\frac{1}{2}$$

D>0

$$36\alpha\ ^2-32\alpha\ \left(1-\alpha\ \right)>0$$

$$68\alpha^2-32\alpha\ >0$$

$$\alpha\ \left(68\alpha\ -32\right)>0$$

$$\alpha\ \in\ \left(-\infty\ ,0\right)U\left(\frac{32}{68},\infty\ \right)$$

Taking the intersection of all we get $$\alpha\ \in\ \left(\frac{32}{68},\frac{1}{2}\right)$$