How do you solve #x^28x=9# by completing the square?
You take half the
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Solve y = x^2  8x  9 = 0
In this case, you don't need to solve by completing the squares. Use the shortcut. When (a  b + c = 0) > 2 real roots: (1) and (c/a = 9).
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To solve the equation (x^2  8x = 9) by completing the square, follow these steps:

Move the constant term to the other side of the equation: (x^2  8x  9 = 0)

To complete the square, take half of the coefficient of (x) (which is (8/2 = 4)) and square it (((4)^2 = 16)). Add this value to both sides of the equation: (x^2  8x + 16 = 9 + 16)

Factor the left side as a perfect square: ((x  4)^2 = 25)

Take the square root of both sides: (x  4 = \pm 5)

Solve for (x): (x = 4 \pm 5)

Therefore, the solutions are (x = 9) and (x = 1).
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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