Certainly! Let's tackle the problem step-by-step.
### 1. Rewrite the Equation by Completing the Square
Given the equation:
[tex]\[ 6x + 55 = x^2 \][/tex]
First, rearrange it to the standard quadratic form:
[tex]\[ x^2 - 6x - 55 = 0 \][/tex]
To complete the square, we follow these steps:
a) Take the coefficient of the linear term (here, -6), halve it, and square the result:
[tex]\[ \left(\frac{-6}{2}\right)^2 = (-3)^2 = 9 \][/tex]
b) Add and subtract this square inside the equation:
[tex]\[ x^2 - 6x + 9 - 9 - 55 = 0 \][/tex]
c) Rearrange to group the perfect square trinomial:
[tex]\[ (x - 3)^2 - 9 - 55 = 0 \][/tex]
d) Simplify:
[tex]\[ (x - 3)^2 - 64 = 0 \][/tex]
e) Rewrite it into the completed square form:
[tex]\[ (x - 3)^2 = 64 \][/tex]
So, the equation in completed square form is:
[tex]\[ (x - 3)^2 = 64 \][/tex]
### 2. Solve the Equation
To find the solutions to the equation [tex]\((x - 3)^2 = 64\)[/tex], we need to take the square root of both sides:
[tex]\[ x - 3 = \pm \sqrt{64} \][/tex]
[tex]\[ x - 3 = \pm 8 \][/tex]
This gives us two possible solutions:
[tex]\[ x - 3 = 8 \Rightarrow x = 8 + 3 \Rightarrow x = 11 \][/tex]
[tex]\[ x - 3 = -8 \Rightarrow x = -8 + 3 \Rightarrow x = -5 \][/tex]
Thus, the solutions to the equation are:
[tex]\[ x = -5 \text{ and } x = 11 \][/tex]
Comparing these solutions with the given choices, they match the solutions in the format of:
(B) [tex]\( x = 3 \pm 8 \)[/tex]
where [tex]\(-5\)[/tex] corresponds to [tex]\(3 - 8\)[/tex] and [tex]\(11\)[/tex] corresponds to [tex]\(3 + 8\)[/tex].
Therefore, the correct answer is:
(B) [tex]\( x = 3 \pm 8 \)[/tex]
