To solve the quadratic equation \( x^2 + 2x - 8 = 0 \) by completing the square, follow these steps:
-
Move the constant to the right side of the equation: \[ x^2 + 2x = 8 \]
-
To complete the square, take half of the coefficient of \( x \) (which is \( 2 \)), square it, and add it to both sides. Half of \( 2 \) is \( 1 \), and \( 1^2 = 1 \): \[ x^2 + 2x + 1 = 8 + 1 \] \[ x^2 + 2x + 1 = 9 \]
-
The left side can now be factored: \[ (x + 1)^2 = 9 \]
-
Now, take the square root of both sides: \[ x + 1 = 3 \quad \text{or} \quad x + 1 = -3 \]
-
Solve for \( x \): \[ x + 1 = 3 \quad \Rightarrow \quad x = 2 \] \[ x + 1 = -3 \quad \Rightarrow \quad x = -4 \]
Thus, the solutions are: \[ x = 2 \quad \text{and} \quad x = -4 \]
So, the correct answer is: x equals 2, x equals negative 4.