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22 April, 23:53

Brandon found that a solution of x = 1 was extraneous for his function. What would be a possible denominator in Brandon's function?

2x + 4

x2 + 3x + 9

3x - 11

x2 + 3x - 4

+3
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  1. 23 April, 00:44
    0
    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.
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