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Question:

Find the set S that denotes the set of all values of 'α' for which the roots of the equation (1 - α)x² - 6αx + 8α = 0 is greater than 2.

Explanation:

αf(x) = (1 - α)x² - 6αx + 8α = 0

Now roots are greater than 2 therefore,

$-\frac{b}{2a}$ > 2

f(2) > 0

D > 0

$-\frac{b}{2a}$ > 0

2(1 - α) > 0

$\frac{\alpha }{\alpha -1}$ < 0

α ∈ (0, 1)

f(2) > 0

(1 − α ) 4 − 12α + 8α > 0
4 − 8α > 0

a < $\frac{1}{2}$

D > 0

36α² - 32α (1 - α) > 0

68α² - 32α > 0

α(68α - 32) > 0

α ∈ (-∞, 0) ∪ $\left(\frac{32}{68},\infty \right)$

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