Quadratic Inequality 8C2Ed8

1. **State the problem:** Solve the quadratic inequality $$x^2 + 3x + 2 < 0$$. 2. **Factor the quadratic expression:** $$x^2 + 3x + 2 = (x + 1)(x + 2)$$ 3. **Set the expression equal to zero to find critical points:** $$(x + 1)(x + 2) = 0$$ This gives the roots: $$x = -1 \quad \text{or} \quad x = -2$$ 4. **Determine intervals to test:** The roots divide the number line into three intervals: $$(-\infty, -2), \quad (-2, -1), \quad (-1, \infty)$$ 5. **Test each interval to see where the inequality holds:** - For $x < -2$, pick $x = -3$: $$(x + 1)(x + 2) = (-3 + 1)(-3 + 2) = (-2)(-1) = 2 > 0$$ - For $-2 < x < -1$, pick $x = -1.5$: $$(x + 1)(x + 2) = (-1.5 + 1)(-1.5 + 2) = (-0.5)(0.5) = -0.25 < 0$$ - For $x > -1$, pick $x = 0$: $$(x + 1)(x + 2) = (0 + 1)(0 + 2) = 1 \times 2 = 2 > 0$$ 6. **Conclusion:** The inequality $$x^2 + 3x + 2 < 0$$ holds true only in the interval: $$-2 < x < -1$$ **Final answer:** $$x > -2 \quad \text{and} \quad x < -1$$