Tricky polynomial root probability problem

Reply #2

db101 — Sun, 12 Feb 2023 06:18:07 GMT

Well that was the last of my set of hard problems. Nice solutions.

Reply #1

cottoneyedjoe — Sat, 11 Feb 2023 18:12:56 GMT

This really wasn't as bad as it looks. I got:

1.) 1/12 = 8.333%

2.) 4/(15*sqrt(3)) = 15.396%

3.) 0

Explanation for 1.) The polynomial x^2 + Ax + B has two real roots when A^2 - 4B = 0, which comes from the discriminant part of the quadratic formula. If you look at the graph of the function B = (A^2)/4 in the A-B-plane over the square -1 A 0 and 0 B 1, the region where B (A^2)/4 has an area of 1/12, which you can get by integrating.

Explanation for 2.) The polynomial x^3... [ More ]

Tricky polynomial root probability problem

db101 — Fri, 10 Feb 2023 06:24:53 GMT

This one is a stew of polynomial roots, probability, and calculus, but it's not as bad as it looks.

Suppose A is a continuous random variable distributed uniformly over (-1, 0) and B is a continuous random variable distributed uniformly over (0, 1).

1.) What is the probability the quadratic polynomial x^2 + Ax + B has two real (possibly repeated) roots?

2.) What is the probability the cubic polynomial x^3 + Ax + B has three real (possibly repeated) roots?

3.) More generally, wh... [ More ]