Question

Complete the square to solve the following quadratic equation: x ^ 2 + 2x - 8 = 0 (1 point) x = 2; x = - 4; x = - 2; x = 4; x = 10x = - 8; x = - 10x = 8

Answer 1

To solve the quadratic 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. To complete the square, we take the coefficient of \( x \) (which is 2), divide it by 2, and then square it. This gives us:
\[
\left(\frac{2}{2}\right)^2 = 1
\]

4. Add and subtract this square (1) to the left side of the equation:
\[
x^2 + 2x + 1 - 1 = 8
\]
Simplifying, we rewrite this as:
\[
(x + 1)^2 - 1 = 8
\]

5. Now, add 1 to both sides:
\[
(x + 1)^2 = 9
\]

6. Take the square root of both sides:
\[
x + 1 = 3 \quad \text{or} \quad x + 1 = -3
\]

7. Solve for \( x \):
\[
x = 3 - 1 \quad \text{or} \quad x = -3 - 1
\]
This gives:
\[
x = 2 \quad \text{or} \quad x = -4
\]

Therefore, the solutions to the equation \( x^2 + 2x - 8 = 0 \) are \( x = 2 \) and \( x = -4 \).