which statement about $\sqrt{x} - \sqrt{x-5} = 1$ is true?

* $x = 4$ is a true solution.
* $x = 4$ is an extraneous solution.
* $x = 9$ is a true solution.
* $x = 9$ is an extraneous solution.

Question

which statement about $\sqrt{x} - \sqrt{x-5} = 1$ is true?

* $x = 4$ is a true solution.
* $x = 4$ is an extraneous solution.
* $x = 9$ is a true solution.
* $x = 9$ is an extraneous solution.

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Radical Equation Solving", "Extraneous Solutions" ], "new_concepts": [], "current_concepts": [ "Radical Equation Solving", "Extraneous Solutions" ] } ## Solve the radical equation $$ \begin{aligned} \sqrt{x} - \sqrt{x-5} &= 1 \ \sqrt{x} - 1 &= \sqrt{x-5} \ (\sqrt{x} - 1)^2 &= x - 5 \ x - 2\sqrt{x} + 1 &= x - 5 \ -2\sqrt{x} &= -6 \ \sqrt{x} &= 3 \ x &= 9 \end{aligned} $$ ## Verify the solution $x = 9$ $$ \begin{aligned} \text{LHS} &= \sqrt{9} - \sqrt{9-5} \ &= 3 - \sqrt{4} \ &= 3 - 2 \ &= 1 \ \text{LHS} &= \text{RHS} \end{aligned} $$ ## Evaluate the alternative value $x = 4$ $$ \begin{aligned} \text{For } x = 4: \quad \sqrt{4-5} &= \sqrt{-1} otin \mathbb{R} \end{aligned} $$

$x = 4$ is a true solution.

$x = 4$ is an extraneous solution.

$x = 9$ is a true solution.

$x = 9$ is an extraneous solution.

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Radical Equation Solving" ] }

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QUESTION IMAGE

Question

Explanation:

Solve the radical equation

Verify the solution \(x = 9\)

Evaluate the alternative value \(x = 4\)

Answer: