a student stated that there are two solutions to the equation $\sqrt{3x + 7} = x - 1$.

they are $x = -1$ and $x = 6$.

do you agree with the student? explain why or why not.

- i do not agree with the student because there is no real solution.
- checking the solutions in the original equation verifies $x = 6$; however, $x = -1$ is extraneous.
- checking the solutions in the original equation shows that 6 is an extraneous solution.
- i agree with the student because both $x = -1$ and $x = 6$ satisfy the original equation.

Question

a student stated that there are two solutions to the equation $\sqrt{3x + 7} = x - 1$.

they are $x = -1$ and $x = 6$.

do you agree with the student? explain why or why not.

- i do not agree with the student because there is no real solution.
- checking the solutions in the original equation verifies $x = 6$; however, $x = -1$ is extraneous.
- checking the solutions in the original equation shows that 6 is an extraneous solution.
- i agree with the student because both $x = -1$ and $x = 6$ satisfy the original equation.

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Solving Radical Equations", "Extraneous Solutions" ], "new_concepts": [], "current_concepts": [ "Solving Radical Equations", "Extraneous Solutions", "Equation Verification" ] } ## Solve the radical equation Using the *Solving Radical Equations* knowledge point $$ \begin{aligned} \sqrt{3x + 7} &= x - 1 \ 3x + 7 &= (x - 1)^2 \ 3x + 7 &= x^2 - 2x + 1 \ x^2 - 5x - 6 &= 0 \ (x - 6)(x + 1) &= 0 \ x = 6 \quad &\text{or} \quad x = -1 \end{aligned} $$ ## Verify the solutions Using the *Extraneous Solutions* and *Equation Verification* knowledge points $$ \begin{aligned} \text{For } x = 6: \quad &\sqrt{3(6) + 7} = \sqrt{25} = 5, \quad 6 - 1 = 5 \quad (\text{Valid}) \ \text{For } x = -1: \quad &\sqrt{3(-1) + 7} = \sqrt{4} = 2, \quad -1 - 1 = -2 \quad (2 eq -2, \text{ Extraneous}) \end{aligned} $$

I do not agree with the student because there is no real solution.

Checking the solutions in the original equation verifies $x = 6$; however, $x = -1$ is extraneous.

Checking the solutions in the original equation shows that 6 is an extraneous solution.

I agree with the student because both $x = -1$ and $x = 6$ satisfy the original equation.

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

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QUESTION IMAGE

Question

Explanation:

Solve the radical equation

Verify the solutions

Answer: