what is the solution set of the quadratic inequality $6x^2 + 1 \le 0$?

$\bigcirc$ $\{x \mid x \in \mathbb{r}\}$

$\bigcirc$ $\{x \mid x = -\frac{1}{6}\}$

$\bigcirc$ $\varnothing$

$\bigcirc$ $\{x \mid x = \frac{1}{6}\}$

Question

what is the solution set of the quadratic inequality $6x^2 + 1 \le 0$?

$\bigcirc$ $\{x \mid x \in \mathbb{r}\}$

$\bigcirc$ $\{x \mid x = -\frac{1}{6}\}$

$\bigcirc$ $\varnothing$

$\bigcirc$ $\{x \mid x = \frac{1}{6}\}$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Quadratic Inequalities" ], "new_concepts": [], "current_concepts": [ "Quadratic Inequalities" ] } ## Analyze the inequality $$ 6x^2 + 1 \le 0 $$ ## Evaluate the range of the quadratic expression $$ \begin{aligned} x^2 &\ge 0 \quad \forall x \in \mathbb{R} \ 6x^2 &\ge 0 \ 6x^2 + 1 &\ge 1 \end{aligned} $$ ## Determine the solution set $$ \begin{aligned} &6x^2 + 1 \ge 1 > 0 \quad \forall x \in \mathbb{R} \ &\text{Therefore, } 6x^2 + 1 \le 0 \text{ has no real solutions.} \ &\text{Solution set} = \varnothing \end{aligned} $$

(A) $\{x \mid x \in \mathbb{R}\}$

(B) $\{x \mid x = -\frac{1}{6}\}$

(C) $\varnothing$

(D) $\{x \mid x = \frac{1}{6}\}$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Quadratic Inequalities" ] }

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QUESTION IMAGE

Question

Explanation:

Analyze the inequality

Evaluate the range of the quadratic expression

Determine the solution set

Answer: