determine the solution for $x^2 + 36 12x$ $\bigcirc$ $\{x \mid x = 6\}$ $\bigcirc$ $\{x \mid -6

Question

determine the solution for $x^2 + 36 > 12x$

$\bigcirc$ $\{x \mid x = 6\}$
$\bigcirc$ $\{x \mid -6 < x < 6\}$
$\bigcirc$ $\{x \mid x \in \mathbb{r}\}$
$\bigcirc$ $\{x \mid x \in \mathbb{r} \text{ and } x \
eq 6\}$

the solution for $x^2 + 2x + 8 \le 0$ is
all real numbers
the empty set
$x = 2$ or $x = 4$
$x = -2$ or $x = 4$

Accepted Answer

{ "quality": "clear", "question_count": 2, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Function Inequalities" ], "new_concepts": [ "Quadratic Inequalities" ], "current_concepts": [ "Function Inequalities", "Quadratic Inequalities" ] } ## Solve the first quadratic inequality $$ \begin{aligned} x^2 + 36 &> 12x \ x^2 - 12x + 36 &> 0 \ (x - 6)^2 &> 0 \ x & eq 6 \ \{x \mid x \in \mathbb{R} \text{ and } x eq 6\} \end{aligned} $$ ## Solve the second quadratic inequality $$ \begin{aligned} x^2 + 2x + 8 &\le 0 \ (x^2 + 2x + 1) + 7 &\le 0 \ (x + 1)^2 + 7 &\le 0 \ (x + 1)^2 &\le -7 \end{aligned} $$ Since $(x+1)^2 \ge 0$ for all real $x$, there are no real solutions. The solution set is the empty set. ## Question 1 {x | x = 6}

{x | -6 < x < 6}

{x | x \in \mathbb{R}}

{x | x \in \mathbb{R} \text{ and } x eq 6} ## Question 2 The solution for $x^2 + 2x + 8 \le 0$ is the empty set { "subject": "Mathematics", "question_type": "Multi-part", "knowledge_point": [ "Mathematics", "Algebra", "Quadratic Inequalities" ] }

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QUESTION IMAGE

Question

Explanation:

Solve the first quadratic inequality

Solve the second quadratic inequality

Answer:

Question 1

Question 2