What is the probability of $f(x) = x^2 - 4x + 2^n$ having 2 zeroes, where $n$ is any real number between $0$ and $6$?

Question

My friend asked about this for his homework. His problem was only for integers, but I'm really interested to know how you would solve this. As far as I can tell, the one restriction is needing a function to determine how many zeroes $f(x)$ will have for any given value of $n$. We can call this function $g(n)$, but I have no idea how it would work. Part of me thinks you'd have to manually check on a graph and make it a piecewise function, but I really want to solve this algebraically.

Answer 1

The function has two zeros if and only if its discriminant is positive, and the discriminant is

$$(-4)^2 - 4 \cdot 1 \cdot 2^n = 16 - 4 \cdot 2^n.$$

So there are two zeros if $n < 2$, one zero if $n = 2$, and none otherwise.

Answer 2

You always have two solutions and this is independent of n, hence the probability is 1. It is of course different if x is constrained to the real numbers.