To solve the quadratic equation \( x^2 + 2x - 8 = 0 \) by completing the square, we can follow these steps:
-
Move the constant to the other side: \[ x^2 + 2x = 8 \]
-
Complete the square:
- Take the coefficient of \( x \) (which is 2), divide it by 2 to get 1, and then square it to get \( 1^2 = 1 \).
- Add this square to both sides of the equation: \[ x^2 + 2x + 1 = 8 + 1 \] \[ x^2 + 2x + 1 = 9 \]
-
Rewrite the left side as a squared term: \[ (x + 1)^2 = 9 \]
-
Take the square root of both sides: \[ x + 1 = \pm 3 \]
-
Solve for \( x \):
- For \( x + 1 = 3 \): \[ x = 3 - 1 = 2 \]
- For \( x + 1 = -3 \): \[ x = -3 - 1 = -4 \]
Thus, the solutions to the equation \( x^2 + 2x - 8 = 0 \) are: \[ x = 2 \quad \text{and} \quad x = -4 \]