GATE CS 2025 Set 2 Q54 — Probability: Random Variables and Distributions

A quadratic polynomial $(x-\alpha)(x-\beta)$ is square invariant if and only if:

Comparing the coefficients on both sides, we get two equations:

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

This gives two cases based on the second equation:

Case 1: $\alpha\beta = 0$
Without loss of generality, let $\alpha = 0$. Substituting this into the first equation gives: $$0 + \beta = 0 + \beta^2 \implies \beta(\beta - 1) = 0 \implies \beta \in \{0, 1\}$$ This yields two distinct polynomials:

• For $\alpha = 0, \beta = 0$: $P_1(x) = x^2$ (roots are equal)
• For $\alpha = 0, \beta = 1$: $P_2(x) = x(x-1)$ (roots are distinct)

Case 2: $\alpha\beta = 1$
Let $t = \alpha + \beta$. Since $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$, the first equation becomes: $$t = t^2 - 2 \implies t^2 - t - 2 = 0 \implies (t-2)(t+1) = 0$$

• If $t = 2$: Since $\alpha+\beta = 2$ and $\alpha\beta = 1$, the polynomial is $P_3(x) = x^2 - 2x + 1 = (x-1)^2$ (roots are equal: $1, 1$).
• If $t = -1$: Since $\alpha+\beta = -1$ and $\alpha\beta = 1$, the polynomial is $P_4(x) = x^2 + x + 1$ (roots are distinct: $\omega, \omega^2$).

Thus, there are exactly $4$ distinct square invariant quadratic polynomials. Among these, $2$ polynomials ($P_1(x)$ and $P_3(x)$) have equal roots.

The probability that the roots of the chosen polynomial are equal is: $$P = \frac{2}{4} = 0.5$$