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)
Answer
0.5
Explanation
Square invariance means the roots \(\{\alpha, \beta\}\) and \(\{\alpha^{2}, \beta^{2}\}\) are the same pair. There are two ways to match them:
- \(\alpha^{2} = \alpha\) and \(\beta^{2} = \beta\): each root is 0 or 1, giving \(x^{2}\), \(x(x - 1)\) and \((x - 1)^{2}\).
- \(\alpha^{2} = \beta\) and \(\beta^{2} = \alpha\): then \(\alpha^{4} = \alpha\), so \(\alpha \in \{0, 1, \omega, \omega^{2}\}\), where \(\omega\) is a primitive cube root of unity. The only new polynomial is \((x - \omega)(x - \omega^{2}) = x^{2} + x + 1\).
So there are exactly 4 square invariant quadratics, and 2 of them, \(x^{2}\) and \((x - 1)^{2}\), have equal roots. The probability is \(2/4 = 0.5\).
Answer: 0.5