To solve the quadratic equation $2x^2 - 7x + 3 = 0$ by completing the square, let's go through the following steps:

Step 1: Divide by the Leading Coefficient

First, divide the entire equation by 2 to make the coefficient of $x^2$ equal to 1.

$2x^2 - 7x + 3 = 0 \implies x^2 - \frac{7}{2}x + \frac{3}{2} = 0$

Step 2: Move the Constant Term to the Right Side

$x^2 - \frac{7}{2}x = -\frac{3}{2}$

Step 3: Complete the Square

To complete the square, take half of the coefficient of $x$, square it, and add it to both sides. The coefficient of $x$ is $-\frac{7}{2}$. Half of this is $-\frac{7}{4}$, and squaring it gives $\left(-\frac{7}{4}\right)^2 = \frac{49}{16}$.

Add $\frac{49}{16}$ to both sides:

$x^2 - \frac{7}{2}x + \frac{49}{16} = -\frac{3}{2} + \frac{49}{16}$

Step 4: Simplify the Right Side

Convert $-\frac{3}{2}$ to a fraction with a denominator of 16:

$-\frac{3}{2} = -\frac{24}{16}$

So:

$-\frac{24}{16} + \frac{49}{16} = \frac{25}{16}$

Now, the equation looks like:

$\left(x - \frac{7}{4}\right)^2 = \frac{25}{16}$

Step 5: Take the Square Root of Both Sides

$x - \frac{7}{4} = \pm \frac{5}{4}$

This gives us two equations:

1. $x - \frac{7}{4} = \frac{5}{4}$
2. $x - \frac{7}{4} = -\frac{5}{4}$

Step 6: Solve for $x$

1. For $x - \frac{7}{4} = \frac{5}{4}$:

$x = \frac{7}{4} + \frac{5}{4} = \frac{12}{4} = 3$

2. For $x - \frac{7}{4} = -\frac{5}{4}$:

$x = \frac{7}{4} - \frac{5}{4} = \frac{2}{4} = \frac{1}{2}$

Solution

The solutions are $x = 3$ and $x = \frac{1}{2}$. The correct answer is:

$\boxed{x = 3, \frac{1}{2}}$

This matches the fourth option in the problem.

Do you have any questions or would you like more details on any part of the solution?

Here are 5 related questions to expand your understanding:

1. How do you solve a quadratic equation using the quadratic formula?
2. Why is it necessary to complete the square when solving quadratics?
3. How would the process change if the equation had no real roots?
4. What are some real-world applications of solving quadratic equations?
5. Can you complete the square when the quadratic term coefficient is not easily divisible?

Tip: When completing the square, always ensure the coefficient of $x^2$ is 1 by dividing through, if necessary.