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)

Question

If probability that $ax^2 + 2\sqrt{2}bx + c > 0 \, \forall x \in R$ where $a, b, c \in {1, 2, 3, 4}$ is $\frac{m}{n}$ where $m$ & $n$ are coprime, then $(m + n)$ is:

Answer 1

A quadratic polynomial $(x-\alpha)(x-\beta)$ is called square invariant if

$(x-\alpha)(x-\beta) = (x-\alpha^2)(x-\beta^2)$

Since two quadratic polynomials are equal if and only if their roots (as an unordered pair) are the same, we must have:

$\{\alpha,\beta\} = \{\alpha^2,\beta^2\}$

Step 1: Possible ways the equality can occur

This happens in either of the following situations:

Each root remains unchanged on squaring: $\alpha=\alpha^2,\; \beta=\beta^2$
The roots are interchanged on squaring: $\alpha=\beta^2,\; \beta=\alpha^2$

Step 2: Roots unchanged on squaring

$z = z^2 \Rightarrow z(z-1)=0$

Hence, possible values are $\alpha,\beta \in \{0,1\}$.

The distinct quadratic polynomials obtained are:

$(x-0)^2$
$(x-1)^2$
$(x-0)(x-1)$

This gives 3 square invariant polynomials.

Step 3: Roots interchange on squaring

$\alpha=\beta^2,\; \beta=\alpha^2 \Rightarrow \alpha^4=\alpha$

$\alpha(\alpha^3-1)=0$

Thus, $\alpha=0$ or $\alpha^3=1$.

Excluding values already counted, this produces one new quadratic polynomial with distinct roots.

Hence, the total number of square invariant quadratic polynomials is:

$3 + 1 = 4$

Step 4: Polynomials with equal roots

Equal roots occur only when:

$(x-0)^2$
$(x-1)^2$

Number of such polynomials = 2

Final Calculation

$\text{Probability} = \dfrac{2}{4} = 0.5$

Answer (rounded to one decimal place):

$\boxed{0.5}$

Answer 2