a student concluded that the solution to the equation $\sqrt{2x+1}+3=0$ is $x=4$.

do you agree? explain why or why not.

Question

a student concluded that the solution to the equation $\sqrt{2x+1}+3=0$ is $x=4$.

do you agree? explain why or why not.

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "explain_then_answer", "tutor", "mastery_adaptive_brevity" ], "needs_drawing": false, "concepts_used": [ "Solving Radical Equations", "Extraneous Solutions", "Principal Square Root Definition" ], "new_concepts": [], "current_concepts": [ "Solving Radical Equations", "Extraneous Solutions", "Principal Square Root Definition" ] } ## Substitute the proposed solution Using the *Solving Radical Equations* knowledge point $$ \begin{aligned} &\text{Substitute } x = 4 \text{ into the left side of the equation:}\ &\sqrt{2(4)+1}+3 = \sqrt{8+1}+3 = \sqrt{9}+3 \end{aligned} $$ ## Evaluate the principal square root Using the *Principal Square Root Definition* knowledge point $$ \begin{aligned} &\sqrt{9} = 3\ &3 + 3 = 6 e 0 \end{aligned} $$ ## Identify the extraneous solution Using the *Extraneous Solutions* knowledge point $$ \begin{aligned} &\sqrt{2x+1} = -3\ &\text{Since } \sqrt{2x+1} \ge 0 \text{ for all real } x, \text{ there are no real solutions.}\ &x = 4 \text{ is an extraneous solution.} \end{aligned} $$ No, because substituting $x = 4$ into the equation yields $6 = 0$, which is false. The value $x = 4$ is an extraneous solution because the principal square root $\sqrt{2x+1}$ cannot be negative, meaning the equation $\sqrt{2x+1} = -3$ has no real solutions. { "subject": "Mathematics", "question_type": "Free Response", "knowledge_point": [ "Mathematics", "Algebra", "Solving Radical Equations" ] }

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QUESTION IMAGE

Question

Explanation:

Substitute the proposed solution

Evaluate the principal square root

Identify the extraneous solution

Answer: