1. **State the problem:** We have a curve given by $$y = x^2 + 2cx + 4$$ and a line given by $$y = 4x + c$$, where $$c$$ is a constant. We want to find the values of $$c$$ for which the curve and line intersect at two distinct points. 2. **Set the equations equal to find intersection points:** At intersection points, the $$y$$ values are equal, so: $$x^2 + 2cx + 4 = 4x + c$$ 3. **Rearrange to form a quadratic equation in $$x$$:** $$x^2 + 2cx + 4 - 4x - c = 0$$ $$x^2 + (2c - 4)x + (4 - c) = 0$$ 4. **Recall the condition for two distinct real roots:** For a quadratic equation $$ax^2 + bx + c = 0$$, two distinct real roots exist if the discriminant $$\Delta = b^2 - 4ac > 0$$. 5. **Calculate the discriminant for our quadratic:** Here, $$a = 1$$, $$b = 2c - 4$$, and $$c = 4 - c$$ (to avoid confusion, denote constant term as $$k = 4 - c$$). $$\Delta = (2c - 4)^2 - 4 \times 1 \times (4 - c)$$ 6. **Simplify the discriminant:** $$\Delta = (2c - 4)^2 - 4(4 - c) = (2c - 4)^2 - 16 + 4c$$ Expand $$ (2c - 4)^2 $$: $$ (2c - 4)^2 = 4c^2 - 16c + 16 $$ So, $$\Delta = 4c^2 - 16c + 16 - 16 + 4c = 4c^2 - 12c$$ 7. **Set the discriminant greater than zero for two distinct points:** $$4c^2 - 12c > 0$$ Divide both sides by 4 (showing cancellation): $$\cancel{4}c^2 - \cancel{4}3c > 0$$ $$c^2 - 3c > 0$$ 8. **Factor the inequality:** $$c(c - 3) > 0$$ 9. **Analyze the inequality:** The product $$c(c - 3)$$ is positive when both factors are positive or both are negative. - Case 1: $$c > 0$$ and $$c - 3 > 0 \Rightarrow c > 3$$ - Case 2: $$c < 0$$ and $$c - 3 < 0 \Rightarrow c < 0$$ 10. **Final solution:** The values of $$c$$ for which the curve and line intersect at two distinct points are: $$c < 0 \quad \text{or} \quad c > 3$$ **Answer:** $$\boxed{c < 0 \text{ or } c > 3}$$