study the steps shown to solve the given equation.
$\sqrt{30 - 2x} = x - 3$
$30 - 2x = x^2 - 6x + 9$
$0 = x^2 - 4x - 21$
$0 = (x + 3)(x - 7)$
based on the above work, possible solutions of the equation are $-3$ and $7$.
check the answers in the original equation. the extraneous solution is $x = $

Question

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "blank_filling_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" ] } ## Identify the potential solutions Using the *Solving Radical Equations* knowledge point $$ \begin{aligned} &\sqrt{30-2x} = x-3 \ &30-2x = (x-3)^2 = x^2-6x+9 \ &x^2-4x-21 = 0 \ &(x+3)(x-7) = 0 \ &x = -3 \quad \text{or} \quad x = 7 \end{aligned} $$ ## Verify the potential solutions Using the *Extraneous Solutions* knowledge point $$ \begin{aligned} &\text{For } x = -3:\ &\text{LHS} = \sqrt{30-2(-3)} = \sqrt{36} = 6 \ &\text{RHS} = -3-3 = -6 \ &\text{LHS} e \text{RHS} \implies x = -3 \text{ is extraneous.} \ \ &\text{For } x = 7:\ &\text{LHS} = \sqrt{30-2(7)} = \sqrt{16} = 4 \ &\text{RHS} = 7-3 = 4 \ &\text{LHS} = \text{RHS} \implies x = 7 \text{ is a valid solution.} \end{aligned} $$ Check the answers in the original equation. The extraneous solution is $x =$ -3 { "subject": "Mathematics", "question_type": "Fill-in-the-blank", "knowledge_point": [ "Mathematics", "Algebra", "Extraneous Solutions" ] }

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Identify the potential solutions

Verify the potential solutions

Answer: