Misleading vieta

$x^2+\dfrac{5}{x-3} = 9 +\dfrac{5}{x-3}$

This is the same as asking:

$x^2 =9 (x\neq 3)$

$\implies \boxed{x=-3}$

$\text{Only one root, so }\boxed{\text{Sum of roots } =x=-3}$

First of all, since $x-3$ is the denominator, $x$ cannot be 3.

When $x$ is not $3$ , then we get $x^2=9$ , so $x=3$ or $-3$ , however $x$ is not $3$ , so the sum of all real roots is $-3$

$\begin{aligned} x^2 + \dfrac{5}{x-3} &= 9 + \dfrac{5}{x-3} \\ x^2 &= 9 &\blue{[x \ne 3]} \\ x &= 3, -3 \\ x &= \boxed{-3} &\blue{[ \because x \ne 3]} \end{aligned}$