sat-dog

Choice B is correct. Subtracting 4 from both sides of isolates the radical expression on the left side of the equation as follows: . Squaring both sides of yields . This equation can be rewritten as a quadratic equation in standard form: . One way to solve this quadratic equation is to factor the expression by identifying two numbers with a sum of and a product of . These numbers are and 1. So the quadratic equation can be factored as . It follows that 5 and are the solutions to the quadratic equation. However, the solutions must be verified by checking whether 5 and satisfy the original equation, . When , the original equation gives , or , which is false. Therefore, does not satisfy the original equation. When , the original equation gives , or , which is true. Therefore, is the only solution to the original equation, and so the solution set is .

Choices A, C, and D are incorrect because each of these sets contains at least one value that results in a false statement when substituted into the given equation. For instance, in choice D, when 0 is substituted for x into the given equation, the result is , or . This is not a true statement, so 0 is not a solution to the given equation.