To complete the square for the equation \(x^2 + 2x - 8 = 0\), we will first move the constant to the other side:
\[ x^2 + 2x = 8 \]
Next, we take the coefficient of \(x\) (which is \(2\)), divide it by \(2\) to get \(1\), and then square it to get \(1^2 = 1\). We add this square to both sides of the equation:
\[ x^2 + 2x + 1 = 8 + 1 \]
This simplifies to:
\[ (x + 1)^2 = 9 \]
Next, we take the square root of both sides:
\[ x + 1 = \pm 3 \]
Now, we can solve for \(x\):
- \(x + 1 = 3 \implies x = 2\)
- \(x + 1 = -3 \implies x = -4\)
Thus, the solutions to the equation \(x^2 + 2x - 8 = 0\) are:
\[ x = 2 \quad \text{and} \quad x = -4 \]
Therefore, the correct response is:
x=2, x=−4
(x equals 2, x equals negative 4)