1. **State the problem:** Solve the equation $$4x^2 + 12x + 9 = 0$$ graphically by drawing the graph of $$y = 4x^2 + 12x + 9$$ for $$-4 \leq x \leq 2$$. 2. **Formula and rules:** The graph of a quadratic function $$y = ax^2 + bx + c$$ is a parabola. The solutions to the equation $$ax^2 + bx + c = 0$$ are the x-values where the parabola intersects the x-axis (i.e., where $$y=0$$). 3. **Rewrite the function:** $$y = 4x^2 + 12x + 9$$ 4. **Complete the square to find vertex form:** $$y = 4(x^2 + 3x) + 9$$ $$= 4\left(x^2 + 3x + \left(\frac{3}{2}\right)^2 - \left(\frac{3}{2}\right)^2\right) + 9$$ $$= 4\left(\left(x + \frac{3}{2}\right)^2 - \frac{9}{4}\right) + 9$$ $$= 4\left(x + \frac{3}{2}\right)^2 - 9 + 9$$ $$= 4\left(x + \frac{3}{2}\right)^2$$ 5. **Interpretation:** The vertex is at $$x = -\frac{3}{2}$$ and $$y=0$$, so the parabola touches the x-axis at this point. 6. **Solve graphically:** Since the parabola only touches the x-axis at $$x = -\frac{3}{2}$$, the equation $$4x^2 + 12x + 9 = 0$$ has one real root (a repeated root) at $$x = -\frac{3}{2}$$. --- Since the user asked to solve both questions but per GUEST RULE only the first problem is solved, the answer is for the first problem only.