The given equation is:

\[
(2x_n) - 1(x - x_n) + x_n^2 - x_{n-1} = 0
\]

To simplify and solve this equation, follow these steps:

1. **Distribute the terms**:
   Expand the term \(-1(x - x_n)\) to get:
   \[
   -1(x - x_n) = -x + x_n
   \]

   Now, substitute back into the equation:
   \[
   (2x_n) - x + x_n + x_n^2 - x_{n-1} = 0
   \]

2. **Combine like terms**:
   Combine all the terms involving \(x_n\):
   \[
   (2x_n + x_n) + x_n^2 - x - x_{n-1} = 0
   \]
   This simplifies to:
   \[
   3x_n + x_n^2 - x - x_{n-1} = 0
   \]

The simplified equation is:

\[
x_n^2 + 3x_n - x - x_{n-1} = 0
\]

This is a recursive relationship involving \(x_n\), \(x_{n-1}\), and \(x\). To proceed further, you'd need either initial values for \(x_n\) and \(x_{n-1}\), or some specific context about the variables.

Would you like more details on solving this or have any questions?

### Related questions:
1. How would this equation change with different initial conditions for \(x_n\)?
2. Can this equation be solved for \(x\) explicitly?
3. What methods are used to solve recursive relations like this?
4. How does the quadratic term \(x_n^2\) influence the solutions?
5. How would you apply this equation to specific sequences?

### Tip:
For recursive sequences, try to express each term in terms of a simpler base case to recognize patterns or derive general formulas.

Explore how to solve the recursive quadratic equation (2x_n) - 1(x - x_n) + x²_n - x_n-1 = 0, with a step-by-step approach. This problem highlights key concepts in recursive relations and quadratic equations, suitable for students in Grades 11-12.

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