How many real roots does $f''(x)f(x) - (f'(x))^2$ have ?

Let's denote the $5$ distinct real roots of $f(x)$ as $a_i , i\in\{1,2,3,4,5\}$ . Now note that

$\frac{f'(x)}{f(x)} = \sum_{i=1}^5\frac{1}{x-a_i}$

$\left( \frac{f'(x)}{f(x)}\right)' = - \sum_{i=1}^5\frac{1}{(x-a_i)^2}$ .

For $x \ne a_i$ , we can write

$f''(x)f(x)-(f'(x))^2 =(f(x))^2\left( \frac{f'(x)}{f(x)}\right)' = -(f(x))^2 \sum_{i=1}^5\frac{1}{(x-a_i)^2}$

So, we see the above expression is negative for real $x \ne a_i$ . So, the only possible candidates for real roots of $f''(x)f(x)-(f'(x))^2$ are the $a_i$ 's.

Now, $f''(a_i)f(a_i)-(f'(a_i))^2 = 0 \Rightarrow f'(a_i)= 0$ . But then $f(a_i) = f'(a_i) = 0$ meaning $a_i$ is a repeated root contradicting that $f$ has all real distinct roots. Thus the $a_i$ 's can't be roots of $f''(x)f(x)-(f'(x))^2$ either. That is, $f''(x)f(x)-(f'(x))^2$ has no real roots.