which solution to the equation $\frac{3}{a+2} + \frac{2}{a} = \frac{4a-4}{a^2-4}$ is extraneous?

$a = -2$
$a = -2$ and $a = 4$
neither $a = -2$ nor $a = 4$
$a = 4$

Question

which solution to the equation $\frac{3}{a+2} + \frac{2}{a} = \frac{4a-4}{a^2-4}$ is extraneous?

$a = -2$
$a = -2$ and $a = 4$
neither $a = -2$ nor $a = 4$
$a = 4$

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 Rational Equations", "Extraneous Solutions" ], "new_concepts": [], "current_concepts": [ "Solving Rational Equations", "Extraneous Solutions", "Rational Expressions" ] } ## Identify excluded values Using the *Solving Rational Equations* knowledge point $$ \begin{aligned} &\text{Denominators: } a+2, \quad a, \quad a^2-4 = (a-2)(a+2)\ &\text{Excluded values: } a eq -2, \quad a eq 0, \quad a eq 2 \end{aligned} $$ ## Solve the rational equation Using the *Solving Rational Equations* knowledge point $$ \begin{aligned} &\frac{3}{a+2} + \frac{2}{a} = \frac{4a-4}{a^2-4}\ &\text{Multiply by LCD: } a(a-2)(a+2)\ &3a(a-2) + 2(a^2-4) = a(4a-4)\ &3a^2 - 6a + 2a^2 - 8 = 4a^2 - 4a\ &5a^2 - 6a - 8 = 4a^2 - 4a\ &a^2 - 2a - 8 = 0\ &(a-4)(a+2) = 0\ &a = 4 \quad \text{or} \quad a = -2 \end{aligned} $$ ## Determine extraneous solutions Using the *Extraneous Solutions* knowledge point $$ \begin{aligned} &a = -2 \text{ is an excluded value (makes denominator zero)}\ &a = 4 \text{ is a valid solution}\ &\text{Extraneous solution: } a = -2 \end{aligned} $$

(A) $a = -2$

(B) $a = -2$ and $a = 4$

(C) neither $a = -2$ nor $a = 4$

(D) $a = 4$

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

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QUESTION IMAGE

Question

Explanation:

Identify excluded values

Solve the rational equation

Determine extraneous solutions

Answer: