An extraneous solution is obtained when the simplified version of the function drives to a value thas does not belong to the domain of the complete function.
So, if x = 1 is an extraneous solution, it does not belong to the original domain.
if when you replace x = 1 in the expression of the denominator it results in 0, then it is an extraneous solution, because you cannot accept it as a valid solution.
Then we must check which of the expressions listed in the options drive to that possibility.
The promising choice is x^2 + 3x - 4.
Inded, If you substitute x = 1, you get (1)2 + 3(1) - 4 = 1 + 3 - 4 = 4 - 4 = 0.
And the answer is x^2 + 3x - 4.
