1. We are asked to solve the inequality $$|x^2 - 2x - 3| < 3$$. 2. Recall that for any expression $A$, the inequality $|A| < b$ (where $b > 0$) means $$-b < A < b$$. 3. Applying this to our problem, we get: $$-3 < x^2 - 2x - 3 < 3$$ 4. We split this into two inequalities: a) $$x^2 - 2x - 3 > -3$$ b) $$x^2 - 2x - 3 < 3$$ 5. Solve inequality (a): $$x^2 - 2x - 3 > -3$$ Add 3 to both sides: $$x^2 - 2x - 3 + 3 > -3 + 3$$ $$x^2 - 2x > 0$$ Factor: $$x(x - 2) > 0$$ 6. The product $x(x-2) > 0$ means both factors are positive or both are negative. - Case 1: $x > 0$ and $x - 2 > 0$ which implies $x > 2$ - Case 2: $x < 0$ and $x - 2 < 0$ which implies $x < 0$ So solution for (a) is: $$x < 0 \quad \text{or} \quad x > 2$$ 7. Solve inequality (b): $$x^2 - 2x - 3 < 3$$ Subtract 3 from both sides: $$x^2 - 2x - 3 - 3 < 3 - 3$$ $$x^2 - 2x - 6 < 0$$ 8. Solve quadratic inequality: $$x^2 - 2x - 6 < 0$$ Find roots of $x^2 - 2x - 6 = 0$ using quadratic formula: $$x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 24}}{2} = \frac{2 \pm \sqrt{28}}{2} = \frac{2 \pm 2\sqrt{7}}{2} = 1 \pm \sqrt{7}$$ 9. Since the parabola opens upward (coefficient of $x^2$ is positive), the inequality $x^2 - 2x - 6 < 0$ holds between the roots: $$1 - \sqrt{7} < x < 1 + \sqrt{7}$$ 10. Combine solutions from steps 6 and 9. The original inequality requires both (a) and (b) to be true simultaneously: $$\left(x < 0 \text{ or } x > 2\right) \quad \text{and} \quad \left(1 - \sqrt{7} < x < 1 + \sqrt{7}\right)$$ 11. Find intersection: - For $x < 0$ and $1 - \sqrt{7} < x < 1 + \sqrt{7}$: Note $1 - \sqrt{7} \approx 1 - 2.6457 = -1.6457$, so interval is $(-1.6457, 3.6457)$. Intersection with $x < 0$ is $(-1.6457, 0)$. - For $x > 2$ and $1 - \sqrt{7} < x < 1 + \sqrt{7}$: Intersection is $(2, 3.6457)$. 12. Final solution: $$x \in (-1.6457, 0) \cup (2, 3.6457)$$ Or exactly: $$x \in (1 - \sqrt{7}, 0) \cup (2, 1 + \sqrt{7})$$