SAT Mathematics Practice Test 4

The question states that $\sqrt{x - a}=x-4$ and that $a=2$ so substituting 2 for a in the equation yields $\sqrt{x-2}=x-4$ To solve for x, square each side of the equation, which gives $(\sqrt{x-2})^2=(x-4)^2$,or $x-2=(x-4)^2$.Then, expanding $(x-4)^2$ yields $x-2=x^2-8x+16,$ or $0=x^2-9x+18$.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 $\sqrt{3-2}=3-4$,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 $\sqrt{6}-2=6-4$, which yields $\sqrt{4}=2$ or$2=2$ Since this is true, the solution set of $\sqrt{x-2}=x-4$ 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.