1. **State the problem:** Solve the inequality $$9x^2 + 3x - 2 \geq 0$$ and graph the solution set on a real number line. 2. **Formula and rules:** To solve a quadratic inequality, first find the roots of the quadratic equation $$9x^2 + 3x - 2 = 0$$ by using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $$a=9$$, $$b=3$$, and $$c=-2$$. 3. **Calculate the discriminant:** $$\Delta = b^2 - 4ac = 3^2 - 4 \times 9 \times (-2) = 9 + 72 = 81$$ 4. **Find the roots:** $$x = \frac{-3 \pm \sqrt{81}}{2 \times 9} = \frac{-3 \pm 9}{18}$$ 5. **Evaluate each root:** - $$x_1 = \frac{-3 + 9}{18} = \frac{6}{18} = \frac{1}{3}$$ - $$x_2 = \frac{-3 - 9}{18} = \frac{-12}{18} = -\frac{2}{3}$$ 6. **Analyze the inequality:** Since $$a=9 > 0$$, the parabola opens upwards. 7. **Determine intervals:** The quadratic $$9x^2 + 3x - 2$$ is: - $$\geq 0$$ outside the roots (i.e., $$x \leq -\frac{2}{3}$$ or $$x \geq \frac{1}{3}$$) - $$< 0$$ between the roots (i.e., $$-\frac{2}{3} < x < \frac{1}{3}$$) 8. **Final solution:** $$x \in (-\infty, -\frac{2}{3}] \cup [\frac{1}{3}, \infty)$$ 9. **Graph the solution set:** The solution set includes all $$x$$ values less than or equal to $$-\frac{2}{3}$$ and greater than or equal to $$\frac{1}{3}$$ on the real number line.