Answer:
The value of x are -2 and -4
Explanation:
Given the equation
[tex]x^2+6x+8=0[/tex]To solve this by completing the square, first of all, we need to rewrite the equation as
[tex]x^2+2(3x)+8=0\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots.(1)[/tex]Note that:
[tex]\begin{gathered} (x+a)^2=x^2+2ax+b^2\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots(2) \\ \\ (x-b)^2=x^2-2ax+b^2\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\text{.}(3) \end{gathered}[/tex]If we write equation (1) as
[tex]x^2+2(3x)+3^2[/tex]we have:
[tex](x+3)^2[/tex]But
[tex]x^2+2(3x)+3^2=x^2+6x+9[/tex]This is equation (1) with an addition of 1.
Subtracting 1 from this will give us exactly equation (1)
[tex](x+3)^2-1=0[/tex]This is exactly equation (1)
Add 1 to both sides of the equation
[tex](x+3)^2=1[/tex]Take square roots of both sides
[tex]\begin{gathered} \sqrt[]{(x+3)^2}=\pm\sqrt[]{1} \\ \\ x+3=\pm1 \\ \\ \text{Subtract 3 from both sides} \\ x=-3\pm1 \end{gathered}[/tex]Therefore,
x = -3 + 1 = -2
OR
x = -3 - 1 = -4
