Correct : 0.5

For the polynomial to be square invariant:
(x−α)(x−β) = (x−α²)(x−β²)

This means {α, β} = {α², β²} as sets.
Case 1: α² = α and β² = β
α(α − 1) = 0 → α = 0 or α = 1
β(β − 1) = 0 → β = 0 or β = 1
Possible pairs: (0,0), (0,1), (1,0), (1,1)
But (0,1) and (1,0) give the same polynomial, so distinct polynomials: (0,0), (1,1), (0,1)
Case 2: α² = β and β² = α (roots swap)
α² = β and β² = α
Substituting: (α²)² = α → α⁴ = α → α⁴ − α = 0
α(α³ − 1) = 0
α = 0 → β = 0 (already in Case 1)
α³ = 1 → α = 1, ω, ω² (cube roots of unity)
α = 1 → β = 1 (already in Case 1)
α = ω → β = ω², gives polynomial (x−ω)(x−ω²)
α = ω² → β = ω, same polynomial as above
All square invariant quadratic polynomials:
1. (x−0)(x−0) = x² → roots equal (0, 0)
2. (x−1)(x−1) = (x−1)² → roots equal (1, 1)
3. (x−0)(x−1) = x(x−1) → roots unequal (0, 1)
4. (x−ω)(x−ω²) → roots unequal (ω, ω²)
Total square invariant polynomials = 4
Polynomials with equal roots = 2 (cases 1 and 2)
Probability = 2 ÷ 4 = 0.5
The probability that the roots of the chosen polynomial are equal = 0.5