what is the solution of $\sqrt{x+12} = x$?

$x = -3$
$x = 4$
$x = -3$ or $x = 4$
no solution

Question

what is the solution of $\sqrt{x+12} = x$?

$x = -3$
$x = 4$
$x = -3$ or $x = 4$
no solution

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" ] } ## Square both sides of the equation Using the *Solving Radical Equations* knowledge point $$ \begin{aligned} \sqrt{x+12} &= x \ (\sqrt{x+12})^2 &= x^2 \ x+12 &= x^2 \end{aligned} $$ ## Solve the quadratic equation Using the *Solving Radical Equations* knowledge point $$ \begin{aligned} x^2 - x - 12 &= 0 \ (x-4)(x+3) &= 0 \ x = 4 \quad &\text{or} \quad x = -3 \end{aligned} $$ ## Check for extraneous solutions Using the *Extraneous Solutions* knowledge point $$ \begin{aligned} &\text{For } x = 4: \quad \sqrt{4+12} = \sqrt{16} = 4 \quad (\text{Valid}) \ &\text{For } x = -3: \quad \sqrt{-3+12} = \sqrt{9} = 3 e -3 \quad (\text{Extraneous}) \end{aligned} $$

$x = -3$

$x = 4$

$x = -3$ or $x = 4$

no solution

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

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Square both sides of the equation

Solve the quadratic equation

Check for extraneous solutions

Answer: