Thank you for pointing that out. Let's correct the solution with the correct right-hand side of the equation.

Given Equation

$\frac{(2x - 1)^2 - (2x + 1)^2}{4x^2 - 1} = -\frac{4}{3}$

Step 1: Expand the Squared Terms

We have two squared terms in the numerator:

$(2x - 1)^2 \quad \text{and} \quad (2x + 1)^2$

Expanding each of them:

$(2x - 1)^2 = 4x^2 - 4x + 1$

$(2x + 1)^2 = 4x^2 + 4x + 1$

Step 2: Substitute the Expanded Forms Back into the Equation

Substituting the expanded forms:

$\frac{(4x^2 - 4x + 1) - (4x^2 + 4x + 1)}{4x^2 - 1} = -\frac{4}{3}$

Step 3: Simplify the Numerator

Simplifying the numerator:

$(4x^2 - 4x + 1) - (4x^2 + 4x + 1) = 4x^2 - 4x + 1 - 4x^2 - 4x - 1$

$= -8x$

Now the equation becomes:

$\frac{-8x}{4x^2 - 1} = -\frac{4}{3}$

Step 4: Cross-Multiply to Simplify

Cross-multiply to eliminate the fraction:

$-8x \cdot 3 = -4 \cdot (4x^2 - 1)$

Simplifying:

$-24x = -16x^2 + 4$

Step 5: Rearrange the Equation

Rearrange all terms to one side to set the equation to zero:

$16x^2 - 24x + 4 = 0$

Step 6: Factor or Use the Quadratic Formula

The equation is now a quadratic equation:

$16x^2 - 24x + 4 = 0$

Step 7: Divide by 4

Divide the entire equation by 4:

$4x^2 - 6x + 1 = 0$

Step 8: Completing the Square

1. Start by moving the constant term to the other side:

   $4x^2 - 6x = -1$

2. Divide the equation by 4 to simplify:

   $x^2 - \frac{3}{2}x = -\frac{1}{4}$

3. Complete the square by adding and subtracting $\left(\frac{3}{4}\right)^2$:

   $x^2 - \frac{3}{2}x + \frac{9}{16} = \frac{9}{16} - \frac{1}{4}$

4. Simplify the right-hand side:

   $\left(x - \frac{3}{4}\right)^2 = \frac{5}{16}$

5. Take the square root of both sides:

   $x - \frac{3}{4} = \pm \frac{\sqrt{5}}{4}$

6. Solve for $x$:

   $x = \frac{3 \pm \sqrt{5}}{4}$

Thus, the solutions are:

$x = \frac{3 + \sqrt{5}}{4} \quad \text{and} \quad x = \frac{3 - \sqrt{5}}{4}$

Would you like further details on this solution or have any other questions?

Follow-Up Questions

1. What are the implications of the discriminant for the solutions obtained?
2. How does completing the square provide insight into the nature of the roots?
3. Can the quadratic formula give the same result, and how would you derive it?
4. What is the significance of the factorization method for solving quadratic equations?
5. How would the solution change if the right-hand side of the equation were a positive fraction?

Tip

When working with quadratic equations, completing the square is particularly useful for converting the equation into vertex form, which can make the properties of the parabola, such as its vertex, more apparent.