To continue solving this recurrence relation, let's pick up from where you left off.

We have: $7\alpha^2 - 3\alpha - 4 = 0$

This is a quadratic equation in terms of $\alpha$, which we can solve using the quadratic formula: $\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 7$, $b = -3$, and $c = -4$.

Step 1: Substitute the values into the formula

$\alpha = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 7 \cdot (-4)}}{2 \cdot 7}$

$= \frac{3 \pm \sqrt{9 + 112}}{14}$

$= \frac{3 \pm \sqrt{121}}{14}$

$= \frac{3 \pm 11}{14}$

Step 2: Solve for the two possible values of $\alpha$

1. For $\alpha = \frac{3 + 11}{14} = \frac{14}{14} = 1$
2. For $\alpha = \frac{3 - 11}{14} = \frac{-8}{14} = -\frac{4}{7}$

So the roots are $\alpha = 1$ and $\alpha = -\frac{4}{7}$.

Step 3: General Solution of the Recurrence Relation

Since we have a second-order linear recurrence relation with distinct roots, the general solution for $U_n$ is given by: $U_n = A \cdot (1)^n + B \cdot \left(-\frac{4}{7}\right)^n$ or simply, $U_n = A + B \cdot \left(-\frac{4}{7}\right)^n$ where $A$ and $B$ are constants determined by initial conditions (which are not provided in the image).

Would you like me to proceed with hypothetical initial conditions, or do you have specific values to find $A$ and $B$?

Questions for Further Exploration:

1. How would you find the constants $A$ and $B$ given specific initial conditions?
2. What is the behavior of $U_n$ as $n \to \infty$?
3. How does the solution change if one or both roots are complex?
4. Can you derive the recurrence relation directly from a known formula for sequences?
5. How does changing the coefficients in the recurrence relation affect the roots?

Tip:

For recurrence relations, always check for real and distinct, real and repeated, or complex roots, as each case gives a different form for the solution.