To solve the quadratic equation \( x^2 + 2x - 8 = 0 \) by completing the square, follow these steps:
-
Move the constant to the other 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 then add it to both sides. Half of 2 is 1, and \( 1^2 \) is 1: \[ x^2 + 2x + 1 = 8 + 1 \] This simplifies to: \[ (x + 1)^2 = 9 \]
-
Take the square root of both sides: \[ x + 1 = \pm 3 \]
-
Solve for \( x \):
- For \( x + 1 = 3 \): \[ x = 3 - 1 = 2 \]
- For \( x + 1 = -3 \): \[ x = -3 - 1 = -4 \]
Now we have the solutions: \[ x = 2, \quad x = -4 \]
Thus, the correct response is:
x=2, x=−4
(x equals 2, x equals negative 4)