To complete the square for the quadratic equation \( x^2 + 2x - 8 = 0 \), we follow these steps:
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Move the constant term to the right side: \[ x^2 + 2x = 8 \]
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Take the coefficient of \( x \), which is 2, divide it by 2 to get 1, and square it to get 1: \[ (1)^2 = 1 \]
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Add this square to both sides: \[ x^2 + 2x + 1 = 8 + 1 \] \[ x^2 + 2x + 1 = 9 \]
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The left side now factors as: \[ (x + 1)^2 = 9 \]
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Take the square root of both sides: \[ x + 1 = \pm 3 \]
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Solve for \( x \): \[ x + 1 = 3 \quad \Rightarrow \quad x = 2 \] \[ x + 1 = -3 \quad \Rightarrow \quad x = -4 \]
Thus, the solutions are: \[ x = 2 \quad \text{and} \quad x = -4 \]
The correct response from the options provided is: x=2, x=−4