SAT Mathematics Practice Test 4

The question states that $\sqrt{\mathit{x}-\mathit{a}}=\mathit{x}-4$ and that $\mathit{a}=2$ so substituting 2 for a in the equation yields $\sqrt{\mathit{x}-2}=\mathit{x}-4$ To solve for x, square each side of the equation, which gives ${\left(\sqrt{\mathit{x}-2}\right)}^2={\left(\mathit{x}-4\right)}^2$,or $\mathit{x}-2={\left(\mathit{x}-4\right)}^2$.Then, expanding ${\left(\mathit{x}-4\right)}^2$ yields $\mathit{x}-2={\mathit{x}}^2-8\mathit{x}+16,$ or $0={\mathit{x}}^2-9\mathit{x}+18$.Factoring the right-hand side gives $0=\left(\mathit{x}-3\right)\left(\mathit{x}-6\right),$and so $\mathit{x}=3$ or $\mathit{x}=6$. However, for x = 3, the original equation becomes $\sqrt{3-2}=3-4$,which yields $1=-1,$ which is not true. Hence, $\mathit{x}=3$ is an extraneous solution that arose from squaring each side of the equation. For $\mathit{x}=6$, 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 $\sqrt{\mathit{x}-2}=\mathit{x}-4$ is $\left|6\right|$
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.