1. The problem asks to find the set of values $x$ such that both inequalities hold: $$x^2 + x - 2 > 0 \quad \text{and} \quad x^2 - 2x - 3 \geq 0$$ 2. First, solve each inequality separately. 3. For $x^2 + x - 2 > 0$: - Factor the quadratic: $$x^2 + x - 2 = (x + 2)(x - 1)$$ - The roots are $x = -2$ and $x = 1$. - Since the parabola opens upward (coefficient of $x^2$ is positive), the inequality $> 0$ holds when $x < -2$ or $x > 1$. 4. For $x^2 - 2x - 3 \geq 0$: - Factor the quadratic: $$x^2 - 2x - 3 = (x - 3)(x + 1)$$ - The roots are $x = 3$ and $x = -1$. - The parabola opens upward, so $\geq 0$ holds when $x \leq -1$ or $x \geq 3$. 5. The combined solution requires both inequalities to be true simultaneously, so take the intersection: $$\{x | x < -2 \text{ or } x > 1\} \cap \{x | x \leq -1 \text{ or } x \geq 3\}$$ 6. Analyze intervals: - For $x < -2$, since $x \leq -1$ includes $x < -2$, the intersection is $x < -2$. - For $x > 1$, intersect with $x \geq 3$ gives $x \geq 3$. 7. Therefore, the solution set is: $$x < -2 \quad \text{or} \quad x \geq 3$$