A quadratic polynomial (x−α)(x−β) over complex numbers is said to be square invariant if (x−α)(x−β)=(x−α2)(x−β2). 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)

The correct answer is 0.5.

To find the probability that a randomly chosen square-invariant quadratic polynomial has equal roots, we first need to determine the total number of such polynomials and then identify how many of them have equal roots.

Let the quadratic polynomial over complex numbers be:
$$P\left(x\right)=(x-\alpha )(x-\beta )$$
The polynomial is said to be square-invariant if:
$$(x-\alpha )(x���\beta )=(x-{\alpha }^2)(x-{\beta }^2)$$
For these two polynomials to be identical, their sets of roots must be equal. Therefore, the set of roots $\{\alpha ,\beta \}$ must be the same as the set $\{{\alpha }^2,{\beta }^2\}$. This gives rise to two distinct cases.

Case 1: Direct Mapping
In this case, each root maps to its own square:
$$\alpha ={\alpha }^2\text{and}\beta ={\beta }^2$$
Solving these equations:
$$\alpha (\alpha -1)=0\Rightarrow \alpha \in \{0,1\}$$
$$\beta (\beta -1)=0\Rightarrow \beta \in \{0,1\}$$
This yields the following possible polynomials:
1. For $\alpha =0,\beta =0$: $P_1\left(x\right)=x^2$ (equal roots)
2. For $\alpha =0,\beta =1$ (or vice versa): $P_2\left(x\right)=x(x-1)$ (distinct roots)
3. For $\alpha =1,\beta =1$: $P_3\left(x\right)={(}^x$ (equal roots)

Case 2: Cross Mapping
In this case, the roots map to each other's squares:
$$\alpha ={\beta }^2\text{and}\beta ={\alpha }^2$$
Substituting the second equation into the first gives:
$$\alpha ={\left({\alpha }^2\right)}^2={\alpha }^4\Rightarrow {\alpha }^4-\alpha =0\Rightarrow \alpha ({\alpha }^3-1)=0$$
Thus, $\alpha \in \{0,1,\omega ,{\omega }^2\}$, where $\omega$ and ${\omega }^2$ are the complex cube roots of unity.
Evaluating each option for $\alpha$ and finding $\beta ={\alpha }^2$:
- If $\alpha =0$, then $\beta =0$ (corresponding to $P_1\left(x\right)$).
- If $\alpha =1$, then $\beta =1$ (corresponding to $P_3\left(x\right)$).
- If $\alpha =\omega$, then $\beta ={\omega }^2$. This produces:
$$P_4\left(x\right)=(x-\omega )(x-{\omega }^2)=x^2+x+1$$ (distinct roots)
- If $\alpha ={\omega }^2,\beta ={\left({\omega }^2\right)}^2=\omega$, which results in the same polynomial $P_4\left(x\right)$.

Summary of Unique Polynomials:
We have a total of 4 unique square-invariant polynomials:
1. $P_1\left(x\right)=x^2$ (roots: 0, 0; equal)
2. $P_2\left(x\right)=x(x-1)$ (roots: 0, 1; distinct)
3. $P_3\left(x\right)={(x-1)}^2$ (roots: 1, 1; equal)
4. $P_4\left(x\right)=x^2+x+1$ (roots: $\omega ,{\omega }^2$; distinct)

Probability Calculation:
The number of polynomials with equal roots is 2 (specifically, $P_1$ and $P_3$).
The total number of square-invariant polynomials is 4.
Therefore, the probability is:
$$\text{Probability}=\frac{2}{4}=0.5$$