The solution set of

is
a) {-1, 2}
b) {2}
c) {0, 1}
d) {1, -2}

  B)  {2}

Step-by-step explanation:

Subtracting the right side, we get ...

The quadratic has solutions x={-1, 2}, but x=-1 makes the original equation undefined. It is extraneous. The only valid solution is x = 2.

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Comment on this solution method

Often, subtracting one side of a rational equation so you get an equation of the form ( )/( ) = 0 will let you cancel common factors from numerator and denominator. This can help avoid any extraneous solutions. (You may still get an extraneous solution if a denominator factor appears at a higher power in the numerator. x³/x² = 0 reduces to x=0, but that is still extraneous.)