To solve the quadratic equation \( x^2 + 2x - 8 = 0 \) by completing the square, follow these steps:
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Move the constant to the right side of the equation:
\[ x^2 + 2x = 8 \]
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Identify the coefficient of \( x \) (which is 2), take half of it (which is 1), and square it (which is 1). We will add and subtract this square on the left side of the equation:
\[ x^2 + 2x + 1 - 1 = 8 \]
This simplifies to:
\[ (x + 1)^2 - 1 = 8 \]
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Rearranging gives:
\[ (x + 1)^2 = 9 \]
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Taking the square root of both sides gives:
\[ x + 1 = 3 \quad \text{or} \quad x + 1 = -3 \]
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Solving for \( x \) in both scenarios:
\( x + 1 = 3 \) leads to:
\[ x = 2 \]
\( x + 1 = -3 \) leads to:
\[ x = -4 \]
Thus, the solutions to the equation \( x^2 + 2x - 8 = 0 \) are:
\[ x = 2 \quad \text{and} \quad x = -4 \]
So the correct response is:
x=2, x=−4
x equals 2, x equals negative 4