To solve the quadratic equation $2x^2 - 3x - 1 = 0$ by completing the square, follow these steps:

Step 1: Move the constant term to the right side

$2x^2 - 3x = 1$

Step 2: Divide by the coefficient of $x^2$

We need the coefficient of $x^2$ to be 1. So, divide the entire equation by 2: $x^2 - \frac{3}{2}x = \frac{1}{2}$

Step 3: Complete the square

To complete the square, take half of the coefficient of $x$ (which is $-\frac{3}{2}$), square it, and add it to both sides: $\left( \frac{-3}{2} \cdot \frac{1}{2} \right)^2 = \left( \frac{-3}{4} \right)^2 = \frac{9}{16}$

Now add $\frac{9}{16}$ to both sides: $x^2 - \frac{3}{2}x + \frac{9}{16} = \frac{1}{2} + \frac{9}{16}$

Step 4: Simplify the right side

$\frac{1}{2} = \frac{8}{16}, \text{ so the right side becomes } \frac{8}{16} + \frac{9}{16} = \frac{17}{16}$

Now we have: $\left( x - \frac{3}{4} \right)^2 = \frac{17}{16}$

Step 5: Solve for $x$

Take the square root of both sides: $x - \frac{3}{4} = \pm \sqrt{\frac{17}{16}} = \pm \frac{\sqrt{17}}{4}$

Step 6: Isolate $x$

Now add $\frac{3}{4}$ to both sides: $x = \frac{3}{4} \pm \frac{\sqrt{17}}{4}$

Final Answer:

The two solutions are: $x = \frac{3 + \sqrt{17}}{4} \quad \text{or} \quad x = \frac{3 - \sqrt{17}}{4}$

Would you like more details or have any questions?

Here are 5 related questions:

1. How is completing the square different from factoring?
2. Can you apply completing the square to any quadratic equation?
3. What is the geometric interpretation of completing the square?
4. How does completing the square help derive the quadratic formula?
5. What are other methods for solving quadratic equations?

Tip: Completing the square is particularly useful for converting quadratics into vertex form!