Answer:
Option B is correct
Step-by-step explanation:
Given: A solution x =3 was extraneous for his function.
Extraneous solution defined as a root of a transformed equation that is not a root of the original equation because it was excluded from the domain of the original equation.
The only function from the given options
f(x) = [tex]x^2-2x-3[/tex] has the solution x =3
therefore,
3 is excluded from the domain of the [tex]\frac{x^2-2x-3}{x-3}[/tex] because it would make the denominator of one of the fractions zero and division by zero is not allowed.
Therefore, it cannot be a root of the equation [tex]\frac{x^2-2x-3}{x-3}[/tex]
