Solve the following equation by completing the square: [(2x^2) - 3x - 1 = 0]

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 other side
\[
2x^2 - 3x = 1
\]

### Step 2: Divide through by the coefficient of \(x^2\) (which is 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.

- Half of \(-\frac{3}{2}\) is \(-\frac{3}{4}\)
- Squaring \(-\frac{3}{4}\) gives \(\left( -\frac{3}{4} \right)^2 = \frac{9}{16}\)

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-hand side
\[
x^2 - \frac{3}{2}x + \frac{9}{16} = \frac{8}{16} + \frac{9}{16} = \frac{17}{16}
\]

### Step 5: Rewrite the left-hand side as a perfect square
\[
\left( x - \frac{3}{4} \right)^2 = \frac{17}{16}
\]

### Step 6: Take the square root of both sides
\[
x - \frac{3}{4} = \pm \frac{\sqrt{17}}{4}
\]

### Step 7: Solve for \(x\)
\[
x = \frac{3}{4} \pm \frac{\sqrt{17}}{4}
\]

So, the two solutions are:
\[
x = \frac{3 + \sqrt{17}}{4} \quad \text{or} \quad x = \frac{3 - \sqrt{17}}{4}
\]

These are the solutions to the equation \(2x^2 - 3x - 1 = 0\).

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Would you like more details or have any questions about the steps? 

Here are some related problems:
1. Solve \(x^2 + 6x + 5 = 0\) by completing the square.
2. Solve \(4x^2 - 12x + 7 = 0\) using the quadratic formula.
3. Convert \(x^2 - 5x + 4 = 0\) into vertex form.
4. Find the roots of \(3x^2 + 2x - 1 = 0\) using factoring.
5. Solve \(x^2 - 7x + 10 = 0\) by graphing.

**Tip:** Completing the square is helpful in deriving the quadratic formula!

Learn how to solve the quadratic equation 2x^2 - 3x - 1 = 0 by completing the square. This step-by-step solution covers algebraic techniques, including how to complete the square, and explains the use of key formulas and theorems. Ideal for students in Grades 9-11.

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