Answer:
1. Add -3
2. Add 4
3. Write the trinomial as [tex](x-2)^2[/tex]
4. [tex]\pm 1[/tex]
Step-by-step explanation:
1. Isolate the constant from the equation [tex]x^2-4x+3=0[/tex] by adding -3 to both sides of the equation. Then
[tex]x^2-4x+3+(-3)=0+(-3),\\ \\x^2-4x=-3.[/tex]
2. Add 4 to both sides of equation [tex]x^2-4x=-3:[/tex]
[tex]x^2-4x+4=-3+4,\\ \\x^2-4x+4=1.[/tex]
3. Write the trinomial [tex]x^2-4x+4[/tex] as perfect square:
[tex]x^2-4x+4=(x-2)^2.[/tex]
4. Use the square root property to get
[tex]x-2=1\text{ or }x-2=-1,\\ \\x=3\text{ or }x=1.[/tex]
