question

solve the rational inequality, and write the solution in interval notation.

\\frac{1}{2} + \frac{12}{x^2} \frac{5}{x}\

provide your answer below:

Question

question

solve the rational inequality, and write the solution in interval notation.

\\frac{1}{2} + \frac{12}{x^2} > \frac{5}{x}\

provide your answer below:

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Rational Inequalities", "Interval Notation" ], "new_concepts": [], "current_concepts": [ "Rational Inequalities", "Interval Notation" ] } ## Write the inequality with a single fraction on one side $$ \begin{aligned} &\frac{1}{2} + \frac{12}{x^2} - \frac{5}{x} > 0\ &\frac{x^2 + 24 - 10x}{2x^2} > 0\ &\frac{x^2 - 10x + 24}{2x^2} > 0 \end{aligned} $$ ## Find the critical points $$ \begin{aligned} &\text{Numerator: } x^2 - 10x + 24 = 0 \implies (x-4)(x-6) = 0 \implies x = 4, \quad x = 6\ &\text{Denominator: } 2x^2 = 0 \implies x = 0 \end{aligned} $$ ## Determine the sign of the expression on each interval $$ \begin{aligned} &\text{Intervals: } (-\infty, 0), \quad (0, 4), \quad (4, 6), \quad (6, \infty)\ &\text{For } x \in (-\infty, 0): \frac{(-)(-)}{(+)} > 0 \implies \text{True}\ &\text{For } x \in (0, 4): \frac{(-)(-)}{(+)} > 0 \implies \text{True}\ &\text{For } x \in (4, 6): \frac{(-)(+)}{(+)} < 0 \implies \text{False}\ &\text{For } x \in (6, \infty): \frac{(+)(+)}{(+)} > 0 \implies \text{True}\ &\text{Excluding } x = 0 \text{ (undefined), } x = 4, \text{ and } x = 6 \text{ (strict inequality)}\ &\text{Solution: } (-\infty, 0) \cup (0, 4) \cup (6, \infty) \end{aligned} $$ $$(-\infty, 0) \cup (0, 4) \cup (6, \infty)$$ { "subject": "Mathematics", "question_type": "Free Response", "knowledge_point": [ "Mathematics", "Algebra", "Rational Inequalities" ] }

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Question

Explanation:

Write the inequality with a single fraction on one side

Find the critical points

Determine the sign of the expression on each interval

Answer: