To solve the equation [tex]\(x^2 + 8x = 38\)[/tex] by completing the square, follow these steps:
1. Move the constant term to the other side of the equation:
[tex]\[
x^2 + 8x - 38 = 0
\][/tex]
2. To complete the square, we need to modify the quadratic expression [tex]\(x^2 + 8x\)[/tex] into a perfect square trinomial. For this, we need to find the constant term that allows us to do this. The formula to complete the square is [tex]\( (b/2)^2 \)[/tex]. In this case, [tex]\(b = 8\)[/tex]:
[tex]\[
\left(\frac{8}{2}\right)^2 = 16.
\][/tex]
3. Add and subtract 16 within the equation to maintain balance:
[tex]\[
x^2 + 8x + 16 - 16 = 38
\][/tex]
Simplifying this, we get:
[tex]\[
x^2 + 8x + 16 = 38 + 16
\][/tex]
[tex]\[
x^2 + 8x + 16 = 54
\][/tex]
4. Now, the left-hand side is a perfect square trinomial:
[tex]\[
(x + 4)^2 = 54
\][/tex]
5. Take the square root of both sides of the equation to solve for [tex]\(x\)[/tex]:
[tex]\[
x + 4 = \pm \sqrt{54}
\][/tex]
Simplify the square root:
[tex]\[
x + 4 = \pm 3\sqrt{6}
\][/tex]
6. Isolate [tex]\(x\)[/tex] by moving 4 to the other side:
[tex]\[
x = -4 \pm 3\sqrt{6}
\][/tex]
Therefore, the solutions can be written as:
[tex]\[
\begin{array}{l}
x = -4 + 3\sqrt{6} \\
x = -4 - 3\sqrt{6}
\end{array}
\][/tex]
So, the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are:
[tex]\[
a = -4, \quad b = 3, \quad c = 6.
\][/tex]
