To complete the square for the quadratic equation \( x^2 + 2x - 8 = 0 \), follow these steps:
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Move the constant to the other side: \( x^2 + 2x = 8 \)
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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 squaring it gives \( 1^2 = 1 \).
Add 1 to both sides: \[ x^2 + 2x + 1 = 8 + 1 \]
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Rewrite the left side as a square: \[ (x + 1)^2 = 9 \]
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Take the square root of both sides: \[ x + 1 = 3 \quad \text{or} \quad x + 1 = -3 \]
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Solve for \( x \): From \( x + 1 = 3 \): \[ x = 3 - 1 = 2 \]
From \( x + 1 = -3 \): \[ x = -3 - 1 = -4 \]
The solutions to the equation are \( x = 2 \) and \( x = -4 \).
So the correct answer is: x = 2, x = −4.