Answer:
[tex]\textsf{B)}\quad \begin{array}{|l|l|}\cline{1-2} \textsf{Step 1}&\vphantom{\dfrac12}x^2 + 6x + 8+1 = 0+1\\\cline{1-2} \textsf{Step 2}&\vphantom{\dfrac12}x^2 + 6x + 9 = 1\\\cline{1-2} \textsf{Step 3}&\vphantom{\dfrac12}(x + 3)^2 = 1\\\cline{1-2}\end{array}[/tex]
Step-by-step explanation:
Given quadratic equation:
[tex]x^2 + 6x + 8 = 0[/tex]
To transform the given equation into the form (x - p)² = q, we need to complete the square.
Begin by adding or subtracting a constant term to both sides of the equation, transforming the left side into a perfect square trinomial that can be factored into the square of a binomial.
To create a perfect square trinomial, we need the constant term to be the square of half the coefficient of the x-term. In this case, the coefficient of the x-term is 6, so the square of half of it is 9. Therefore, we need to add 1 to both sides of the equation to ensure the constant on the left side of the equation is equal to 9:
[tex]x^2 + 6x + 8+1 = 0+1[/tex]
[tex]x^2 + 6x + 9 = 1[/tex]
Now we can factor the perfect square trinomial on the left of the equation. A perfect square trinomial a² + 2ab + b² can be factored into (a + b)². So, in the case of x² + 6x + 9, a = x and b = 3. Therefore:
[tex](x + 3)^2 = 1[/tex]
Answer:
[tex]\large \textsf{Read below}[/tex]
Step-by-step explanation:
[tex]\large \text{$ \sf x^2 + 6x + 8 = 0$}[/tex]
[tex]\large \textsf{Step 1}[/tex]
[tex]\large \text{$ \sf x^2 + 6x + 8 + 1 = 0 + 1$}[/tex]
[tex]\large \textsf{Step 2}[/tex]
[tex]\large \text{$ \sf x^2 + 6x + 9 = 1$}[/tex]
[tex]\large \textsf{Step 3}[/tex]
[tex]\large \text{$ \sf (x + 3)^2 = 1$}[/tex]
