How do you solve #x^2 + 6x = 3# using completing the square?
To complete the square you add
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To solve ( x^2 + 6x = 3 ) using completing the square:
- Move the constant term to the other side: ( x^2 + 6x - 3 = 0 ).
- Add and subtract ((\frac{6}{2})^2 = 9) within the expression: ( x^2 + 6x + 9 - 9 - 3 = 0 ).
- Rewrite the expression as a perfect square trinomial and simplify: ( (x + 3)^2 - 12 = 0 ).
- Add 12 to both sides: ( (x + 3)^2 = 12 ).
- Take the square root of both sides: ( x + 3 = \pm \sqrt{12} ).
- Simplify the square root: ( x + 3 = \pm 2\sqrt{3} ).
- Subtract 3 from both sides: ( x = -3 \pm 2\sqrt{3} ).
So, the solutions to the equation ( x^2 + 6x = 3 ) using completing the square are ( x = -3 + 2\sqrt{3} ) and ( x = -3 - 2\sqrt{3} ).
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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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