what is the solution set of the quadratic inequality $x^2 + x - 2 \ge 0$?

$\{x \mid x \le -2 \text{ or } x \ge 1\}$
$\{x \mid x \le -1 \text{ or } x \ge 2\}$
$\{x \mid x \ge -2 \text{ or } x \le 1\}$
$\{x \mid x \ge -1 \text{ or } x \le 2\}$

Question

what is the solution set of the quadratic inequality $x^2 + x - 2 \ge 0$?

$\{x \mid x \le -2 \text{ or } x \ge 1\}$
$\{x \mid x \le -1 \text{ or } x \ge 2\}$
$\{x \mid x \ge -2 \text{ or } x \le 1\}$
$\{x \mid x \ge -1 \text{ or } x \le 2\}$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Polynomial Factoring", "Polynomial Inequalities" ], "new_concepts": [], "current_concepts": [ "Polynomial Factoring", "Polynomial Inequalities", "Sign Analysis Intervals" ] } ## Factor the quadratic expression $$ x^2 + x - 2 = (x + 2)(x - 1) $$ ## Find the critical points $$ (x + 2)(x - 1) = 0 \implies x = -2 \quad \text{or} \quad x = 1 $$ ## Determine the solution intervals $$ \begin{aligned} &\text{For } x \le -2: \quad (x+2 \le 0) \text{ and } (x-1 \le 0) \implies (x+2)(x-1) \ge 0\ &\text{For } -2 < x < 1: \quad (x+2 > 0) \text{ and } (x-1 < 0) \implies (x+2)(x-1) < 0\ &\text{For } x \ge 1: \quad (x+2 \ge 0) \text{ and } (x-1 \ge 0) \implies (x+2)(x-1) \ge 0\ &\text{Thus, } x^2 + x - 2 \ge 0 \implies \{x \mid x \le -2 \text{ or } x \ge 1\} \end{aligned} $$

(A) $\{x \mid x \le -2 \text{ or } x \ge 1\}$

(B) $\{x \mid x \le -1 \text{ or } x \ge 2\}$

(C) $\{x \mid x \ge -2 \text{ or } x \le 1\}$

(D) $\{x \mid x \ge -1 \text{ or } x \le 2\}$

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

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QUESTION IMAGE

Question

Explanation:

Factor the quadratic expression

Find the critical points

Determine the solution intervals

Answer: