To complete the square for the quadratic equation \( x^2 + 2x - 8 = 0 \), follow these steps:
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Move the constant term to the right side: \[ x^2 + 2x = 8 \]
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Take half of the coefficient of \( x \) (which is 2), square it, and add it to both sides. The half of 2 is 1, and squaring it gives \( 1^2 = 1 \): \[ x^2 + 2x + 1 = 8 + 1 \]
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This simplifies to: \[ (x + 1)^2 = 9 \]
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Now, take the square root of both sides: \[ x + 1 = \pm 3 \]
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Solve for \( x \):
- For \( x + 1 = 3 \): \[ x = 3 - 1 = 2 \]
- For \( x + 1 = -3 \): \[ x = -3 - 1 = -4 \]
The solutions to the equation \( x^2 + 2x - 8 = 0 \) are: \[ x = 2 \quad \text{and} \quad x = -4. \]