# How do you solve by completing the square for #x^2-8x+7#?

You have to take half the

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To solve by completing the square for (x^2 - 8x + 7), follow these steps:

- Rewrite the equation as (x^2 - 8x + \left(\frac{-8}{2}\right)^2 - \left(\frac{-8}{2}\right)^2 + 7).
- Simplify inside the parentheses: (x^2 - 8x + 16 - 16 + 7).
- Combine like terms: (x^2 - 8x + 16 - 9).
- Rewrite as a perfect square trinomial: ((x - 4)^2 - 9).
- The solution is ((x - 4)^2 - 9 = 0).
- Add 9 to both sides: ((x - 4)^2 = 9).
- Take the square root of both sides: (x - 4 = \pm 3).
- Solve for (x): (x = 4 \pm 3).
- The solutions are (x = 4 + 3 = 7) and (x = 4 - 3 = 1).

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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.

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