Complete each statement in the steps to solve [tex]$x^2 - 4x + 3 = 0$[/tex] using the process of completing the square.

1. Isolate the constant by [tex]\(\square\)[/tex] both sides of the equation.
2. Add [tex]\(\square\)[/tex] to both sides of [tex]$x^2 - 4x = -3$[/tex] to form a perfect square trinomial while keeping the equation balanced.
3. Write the trinomial [tex]$x^2 - 4x + 4$[/tex] as [tex]\(\square\)[/tex] squared.
4. Use the square root property of equality to get [tex]$x - 2 = \pm \square$[/tex].
5. Isolate the variable to get solutions of 1 and 3.



Answer :

Of course! Let's complete each statement for solving [tex]\( x^2 - 4x + 3 = 0 \)[/tex] using the process of completing the square step by step.

1. Isolate the constant by subtracting 3 from both sides of the equation.
[tex]\[ x^2 - 4x + 3 - 3 = 0 - 3 \implies x^2 - 4x = -3 \][/tex]

2. Add 4 to both sides of [tex]\( x^2 - 4x = -3 \)[/tex] to form a perfect square trinomial while keeping the equation balanced.
[tex]\[ x^2 - 4x + 4 = -3 + 4 \implies x^2 - 4x + 4 = 1 \][/tex]

3. Write the trinomial [tex]\( x^2 - 4x + 4 \)[/tex] as [tex]\( (x-2) \)[/tex] squared.
[tex]\[ x^2 - 4x + 4 = (x - 2)^2 \][/tex]

4. Use the square root property of equality to get [tex]\( x-2 = \pm \)[/tex] 1 [tex]\(\checkmark\)[/tex].
[tex]\[ (x - 2)^2 = 1 \implies x - 2 = \pm 1 \][/tex]

5. Isolate the variable to get solutions of 1 and 3.
[tex]\[ \text{If } x - 2 = 1, \text{ then } x = 2 + 1 = 3 \][/tex]
[tex]\[ \text{If } x - 2 = -1, \text{ then } x = 2 - 1 = 1 \][/tex]

Thus, the solutions to the equation [tex]\( x^2 - 4x + 3 = 0 \)[/tex] are [tex]\( x = 3 \)[/tex] and [tex]\( x = 1 \)[/tex].