Q. Convert the equation \(x^2-4x+3=0\). Complete the square.

We want to solve the equation by completing the square:

First, start with the given quadratic:

\[
x^2 – 4x + 3 = 0
\]

Step 1: Move the constant term to the other side

We isolate the part with \(x\) so that we can complete the square easily.

\[
x^2 – 4x = -3
\]

Step 2: Complete the square for \(x^2 – 4x\)

Recall the rule: to complete the square, take half of the coefficient of \(x\), square it, and add and subtract it appropriately.

Here, the coefficient of \(x\) in \(x^2 – 4x\) is \(-4\).

Compute half of \(-4\):

\[
\frac{-4}{2} = -2
\]

Now square \(-2\):

\[
(-2)^2 = 4
\]

So we will add and subtract \(4\) inside the expression \(x^2 – 4x\).

Rewrite the left side by grouping:

\[
x^2 – 4x + 4 – 4 = -3
\]

Group the first three terms to form a perfect square trinomial:

\[
(x^2 – 4x + 4) – 4 = -3
\]

Recognize the perfect square:

\[
(x – 2)^2 – 4 = -3
\]

Step 3: Add \(4\) to both sides

We want the square term by itself.

\[
(x – 2)^2 = -3 + 4
\]

Compute the right side:

\[
(x – 2)^2 = 1
\]

Step 4: Take the square root of both sides

If \((x – 2)^2 = 1\), then \(x – 2\) can be \(1\) or \(-1\).

\[
x – 2 = 1
\]

and

\[
x – 2 = -1
\]

Step 5: Solve each linear equation

First case:

\[
x – 2 = 1
\]

\[
x = 3
\]

Second case:

\[
x – 2 = -1
\]

\[
x = 1
\]

Final Answer

The solutions are:

\[
x = 1 \quad \text{or} \quad x = 3
\]