which of the following is an extraneous solution of $(45-3x)^{\frac{1}{2}} = x-9$?

$x = -12$
$x = -3$
$x = 3$
$x = 12$

Question

which of the following is an extraneous solution of $(45-3x)^{\frac{1}{2}} = x-9$?

$x = -12$
$x = -3$
$x = 3$
$x = 12$

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", "Quadratic Formula", "Polynomial Factoring" ] } ## Square both sides of the equation Using the **Solving Radical Equations** knowledge point $$ \begin{aligned} (45-3x)^{\frac{1}{2}} &= x-9 \ 45-3x &= (x-9)^2 \ 45-3x &= x^2 - 18x + 81 \end{aligned} $$ ## Solve the quadratic equation Using the **Polynomial Factoring** knowledge point $$ \begin{aligned} x^2 - 15x + 36 &= 0 \ (x-12)(x-3) &= 0 \ x = 12 \quad &\text{or} \quad x = 3 \end{aligned} $$ ## Check for extraneous solutions Using the **Extraneous Solutions** knowledge point $$ \begin{aligned} &\text{For } x = 12:\ &\text{LHS} = (45-3(12))^{\frac{1}{2}} = 9^{\frac{1}{2}} = 3\ &\text{RHS} = 12 - 9 = 3 \quad (\text{Valid})\ \ &\text{For } x = 3:\ &\text{LHS} = (45-3(3))^{\frac{1}{2}} = 36^{\frac{1}{2}} = 6\ &\text{RHS} = 3 - 9 = -6 \quad (\text{Invalid, } 6 eq -6) \end{aligned} $$ Thus, $x = 3$ is the extraneous solution.

x = -12

x = -3

x = 3

x = 12

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

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QUESTION IMAGE

Question

Explanation:

Square both sides of the equation

Solve the quadratic equation

Check for extraneous solutions

Answer: