How do you solve using the completing the square method #x^2+2x-8=0#?
The solutions are:
We add one to both sides in order to write the left-hand side as a perfect square:
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To solve (x^2 + 2x - 8 = 0) using the completing the square method, follow these steps:
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Move the constant term to the other side of the equation: (x^2 + 2x = 8)
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Take half of the coefficient of (x), square it, and add it to both sides of the equation: (x^2 + 2x + (2/2)^2 = 8 + (2/2)^2) (x^2 + 2x + 1 = 8 + 1)
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Rewrite the left side as a perfect square: ((x + 1)^2 = 9)
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Take the square root of both sides: (x + 1 = ±\sqrt{9})
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Solve for (x): (x = -1 ± 3)
So, the solutions are (x = -1 + 3) and (x = -1 - 3), which simplifies to (x = 2) and (x = -4).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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