1. **Solve the first system of inequalities:** Given: $$\begin{cases} x^2 + 2x + 5 > 0 \\ (x - 1)(x^2 - 4) \geq 0 \end{cases}$$ - Step 1: Analyze the first inequality $x^2 + 2x + 5 > 0$. The quadratic $x^2 + 2x + 5$ has discriminant $\Delta = 2^2 - 4 \times 1 \times 5 = 4 - 20 = -16 < 0$, so it has no real roots. Since the leading coefficient is positive, the quadratic is always positive for all real $x$. Therefore, $$x^2 + 2x + 5 > 0 \quad \text{for all real } x.$$ - Step 2: Analyze the second inequality $(x - 1)(x^2 - 4) \geq 0$. Factor $x^2 - 4$ as $(x - 2)(x + 2)$. So the inequality is: $$ (x - 1)(x - 2)(x + 2) \geq 0 $$ - Step 3: Find critical points: $x = -2, 1, 2$. - Step 4: Test intervals determined by these points: - For $x < -2$, pick $x = -3$: $(-3 - 1)(-3 - 2)(-3 + 2) = (-4)(-5)(-1) = -20 < 0$ - For $-2 < x < 1$, pick $x = 0$: $(0 - 1)(0 - 2)(0 + 2) = (-1)(-2)(2) = 4 > 0$ - For $1 < x < 2$, pick $x = 1.5$: $(1.5 - 1)(1.5 - 2)(1.5 + 2) = (0.5)(-0.5)(3.5) = -0.875 < 0$ - For $x > 2$, pick $x = 3$: $(3 - 1)(3 - 2)(3 + 2) = (2)(1)(5) = 10 > 0$ - Step 5: Include points where expression equals zero: $x = -2, 1, 2$. - Step 6: Solution for second inequality: $$[-2, 1] \cup [2, \infty)$$ - Step 7: Since first inequality is always true, solution for first system is: $$\boxed{[-2, 1] \cup [2, \infty)}$$ 2. **Solve the second system of inequalities:** Given: $$\begin{cases} x^2 + 6x + 8 \geq 0 \\ x^2 - 9 > 0 \end{cases}$$ - Step 1: Factor $x^2 + 6x + 8$: $$x^2 + 6x + 8 = (x + 2)(x + 4)$$ - Step 2: Solve $x^2 + 6x + 8 \geq 0$: Critical points: $x = -4, -2$. Test intervals: - $x < -4$, pick $x = -5$: $( -5 + 2)(-5 + 4) = (-3)(-1) = 3 > 0$ - $-4 < x < -2$, pick $x = -3$: $(-3 + 2)(-3 + 4) = (-1)(1) = -1 < 0$ - $x > -2$, pick $x = 0$: $(0 + 2)(0 + 4) = 2 \times 4 = 8 > 0$ Include points where expression equals zero: $x = -4, -2$. Solution: $$(-\infty, -4] \cup [-2, \infty)$$ - Step 3: Solve $x^2 - 9 > 0$: Factor as $(x - 3)(x + 3) > 0$. Critical points: $x = -3, 3$. Test intervals: - $x < -3$, pick $x = -4$: $( -4 - 3)(-4 + 3) = (-7)(-1) = 7 > 0$ - $-3 < x < 3$, pick $x = 0$: $(0 - 3)(0 + 3) = (-3)(3) = -9 < 0$ - $x > 3$, pick $x = 4$: $(4 - 3)(4 + 3) = (1)(7) = 7 > 0$ Solution: $$(-\infty, -3) \cup (3, \infty)$$ - Step 4: Combine both inequalities with AND: $$\left((-\infty, -4] \cup [-2, \infty)\right) \cap \left((-\infty, -3) \cup (3, \infty)\right)$$ - Step 5: Find intersections: - $(-\infty, -4] \cap (-\infty, -3) = (-\infty, -4]$ (since $-4 < -3$) - $[-2, \infty) \cap (-\infty, -3) = \emptyset$ - $(-\infty, -4] \cap (3, \infty) = \emptyset$ - $[-2, \infty) \cap (3, \infty) = (3, \infty)$ - Step 6: Final solution: $$\boxed{(-\infty, -4] \cup (3, \infty)}$$ **Final answers:** - First system solution: $$[-2, 1] \cup [2, \infty)$$ - Second system solution: $$(-\infty, -4] \cup (3, \infty)$$