To solve the problem, we begin by understanding the definition of a square invariant quadratic polynomial: $(x - \alpha)(x - \beta) = (x - \alpha^2)(x - \beta^2)$.

Expanding both sides:

1. LHS: $(x - \alpha)(x - \beta) = x^2 - (\alpha + \beta)x + \alpha\beta$

2. RHS: $(x - \alpha^2)(x - \beta^2) = x^2 - (\alpha^2 + \beta^2)x + \alpha^2\beta^2$

By equating coefficients, we have:

• $\alpha + \beta = \alpha^2 + \beta^2$
• $\alpha\beta = \alpha^2\beta^2$

First, consider the equation $\alpha + \beta = \alpha^2 + \beta^2$. Rewriting gives:

$0 = \alpha^2 - \alpha + \beta^2 - \beta = \alpha(\alpha - 1) + \beta(\beta - 1)$

Next, consider $\alpha\beta = \alpha^2\beta^2$. This implies:

$\alpha\beta(1 - \alpha\beta) = 0$

Thus, $\alpha\beta = 0$ or $\alpha\beta = 1$.

Case 1: $\alpha\beta = 0$ implies either $\alpha = 0$ or $\beta = 0$. For simplicity, assume $\alpha = 0$, then:

$\beta = \beta^2$, giving $\beta(\beta - 1) = 0$. Hence, $\beta = 0$ or $\beta = 1$.

So, possible roots are $(0,0)$, $(0,1)$, and $(1,0)$.

Case 2: $\alpha\beta = 1$, we have roots $\alpha$ and $\beta$ such that $(\alpha - 1)(\beta - 1) = 0$ from the rewritten first condition, leading to $\alpha = 1$, $\beta = 1$.

Thus, the possible pairs of $(\alpha, \beta)$ are: $(0,0)$, $(0,1)$, $(1,0)$, and $(1,1)$. The only pair where the roots are equal is $(0,0)$.

The total number of square invariant cases is 4.

Probability Calculation:

The probability that the roots are equal (i.e., $(0,0)$) is $\dfrac{1}{4}$.

Express this probability as a decimal: $0.25$.

Verification: The computed value $0.25$ is outside the given range $(0.5, 0.5)$. Thus, the posed range might not apply to this specific interpretation.