1. **State the problem:** Solve the equation $$|x| = x^2 + x - 3$$. 2. **Recall the definition of absolute value:** $$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$ 3. **Split the problem into two cases based on the definition:** **Case 1: $x \geq 0$** The equation becomes: $$x = x^2 + x - 3$$ Simplify by subtracting $x$ from both sides: $$0 = x^2 - 3$$ 4. **Solve the quadratic:** $$x^2 = 3$$ $$x = \pm \sqrt{3}$$ Since $x \geq 0$ in this case, only $x = \sqrt{3}$ is valid. 5. **Case 2: $x < 0$** The equation becomes: $$-x = x^2 + x - 3$$ Bring all terms to one side: $$0 = x^2 + x - 3 + x$$ $$0 = x^2 + 2x - 3$$ 6. **Solve the quadratic:** $$x^2 + 2x - 3 = 0$$ Factor or use quadratic formula: $$x = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-3)}}{2} = \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm \sqrt{16}}{2}$$ $$x = \frac{-2 \pm 4}{2}$$ Two solutions: $$x = \frac{-2 + 4}{2} = 1$$ $$x = \frac{-2 - 4}{2} = -3$$ Since $x < 0$ in this case, only $x = -3$ is valid. 7. **Final solutions:** $$x = \sqrt{3} \quad \text{or} \quad x = -3$$ These satisfy the original equation.