A quadratic polynomial $(x - \alpha)(x - \beta)$ over complex numbers is said to be square invariant if $(x - \alpha)(x - \beta) = (x - \alpha^2)(x - \beta^2)$. Suppose from the set of all square invariant quadratic polynomials we choose one at random. The probability that the roots of the chosen polynomial are equal is ________. (rounded off to one decimal place)

A monic quadratic polynomial is uniquely determined by the multiset of its roots. The condition:

means that the multiset of roots $\{\alpha, \beta\}$ must equal the multiset $\{\alpha^2, \beta^2\}$.

There are two cases for multiset equality:

1. Case 1: $\alpha = \alpha^2$ and $\beta = \beta^2$

   • The solutions to $z = z^2$ are $z \in \{0, 1\}$.

   • Possible unordered root pairs (and the corresponding distinct polynomials):

     • $\{\alpha, \beta\} = \{0, 0\} \implies P_1(x) = x^2$ (roots are equal)

     • $\{\alpha, \beta\} = \{1, 1\} \implies P_2(x) = (x-1)^2$ (roots are equal)

     • $\{\alpha, \beta\} = \{0, 1\} \implies P_3(x) = x(x-1) = x^2 - x$ (roots are distinct)

2. Case 2: $\alpha = \beta^2$ and $\beta = \alpha^2$

   • Substituting gives $\alpha = (\alpha^2)^2 = \alpha^4 \implies \alpha(\alpha^3 - 1) = 0$.

   • The solutions are $\alpha \in \{0, 1, \omega, \omega^2\}$, where $\omega, \omega^2$ are the non-real cube roots of unity ($e^{i 2\pi/3}$ and $e^{i 4\pi/3}$).

   • For $\alpha = \omega$, we have $\beta = \omega^2$, giving:

     • $\{\alpha, \beta\} = \{\omega, \omega^2\} \implies P_4(x) = (x-\omega)(x-\omega^2) = x^2 + x + 1$ (roots are distinct)

   • The cases $\alpha = 0$ and $\alpha = 1$ give polynomials already accounted for in Case 1.

Thus, there are a total of 4 distinct square-invariant quadratic polynomials:

1. $P_1(x) = x^2$ (equal roots: $0, 0$)

2. $P_2(x) = (x-1)^2$ (equal roots: $1, 1$)

3. $P_3(x) = x^2 - x$ (distinct roots: $0, 1$)

4. $P_4(x) = x^2 + x + 1$ (distinct roots: $\omega, \omega^2$)

Among these 4 polynomials, exactly 2 have equal roots ($P_1$ and $P_2$).