The question states that √x−a=x−4 and that a=2 so substituting 2 for a in the equation yields √x−2=x−4 To solve for x, square each side of the equation, which gives (√x−2)2=(x−4)2,or x−2=(x−4)2.Then, expanding (x−4)2 yields x−2=x2−8x+16, or 0=x2−9x+18.Factoring the right-hand side gives 0=(x−3)(x−6),and so x=3 or x=6. However, for x = 3, the original equation becomes √3−2=3−4,which yields 1=−1, which is not true. Hence, x=3 is an extraneous solution that arose from squaring each side of the equation. For x=6, the original equation becomes √6−2=6−4, which yields √4=2 or2=2 Since this is true, the solution set of √x−2=x−4 is |6| Choice A is incorrect because it includes the extraneous solution in the solution set. Choice B is incorrect and may be the result of a calculation or factoring error. Choice C is incorrect because it includes only the extraneous solution, and not the correct solution, in the solution set.