When is a function not defined?

The function is not defined for the roots(values of x) of the denominator. Since we want to find the sum of all these x, we know for a quadratic equation, if α and β are the roots of the equation, then $α+β=\frac{-b}{a}.$ Here a=1 and b =-20. Therefore $α+β=\frac{-(-20)}{1}=\boxed{20}$ which is our answer.

A function is defined only when it has a non-zero denominator (we don't care what numerator is) because we have not defined what division by zero is . So , the denominator of Andrew's function is $x^2-20x+91$ , and for the function to be defined , we have:

$x^2-20x+91 \neq 0 \Rightarrow (x-7)(x-13)\neq 0 \Rightarrow x\neq 7,13$

Hence , sum of required numbers $=7+13=\boxed{20}$ .

We can factorize the function which Andrew wrote as follows:

$f(x) = \dfrac{x-13}{x^2-20x+91}= \dfrac{x-13}{(x-7)(x-13)}$

Hence when $x = 13$ , $f(13)=\dfrac{13-13}{(13-7)(13-13)}=\dfrac 00$ becomes undefined.

And similarly, at $x=7$ , $f(7) = \dfrac{7-13}{(7-7)(7-13)}=\dfrac{-6}0$ becomes undefined.

The function is not defined when the $denominator = 0 \\ \Rightarrow x^2 - 20x + 91 = 0 \\ \Rightarrow (x - 13)(x - 7) = 0 \\ \Rightarrow x = 13,7. \\ \therefore Answer = Sum = 13 + 7 = \color{#3D99F6}{20}$ .