To solve the equation \( x^2 + 2x - 8 = 0 \) by completing the square, follow these steps:
1. Start with the original equation:
\[
x^2 + 2x - 8 = 0
\]
2. Move the constant term to the other side of the equation:
\[
x^2 + 2x = 8
\]
3. Complete the square on the left side. To do this, take half of the coefficient of \( x \), square it, and add it to both sides. The coefficient of \( x \) is 2:
\[
\left(\frac{2}{2}\right)^2 = 1
\]
Add 1 to both sides:
\[
x^2 + 2x + 1 = 8 + 1
\]
\[
x^2 + 2x + 1 = 9
\]
4. Now, rewrite the left side as a square:
\[
(x + 1)^2 = 9
\]
5. Take the square root of both sides:
\[
x + 1 = \pm 3
\]
6. Solve for \( x \):
- If \( x + 1 = 3 \), then \( x = 2 \).
- If \( x + 1 = -3 \), then \( x = -4 \).
Thus, the solutions to the equation are:
\[
x = 2 \quad \text{and} \quad x = -4
\]
So the correct answer is:
**x=2, x=−4** (or \( x \) equals 2, \( x \) equals negative 4).