How do you solve using the completing the square method #3x^26x1=0#?
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To solve the quadratic equation (3x^2  6x  1 = 0) using the completing the square method:

Move the constant term to the other side of the equation: (3x^2  6x = 1)

Divide the entire equation by the coefficient of (x^2) (in this case, 3) to make the coefficient of (x^2) equal to 1: (x^2  2x = \frac{1}{3})

To complete the square, take half of the coefficient of (x) (2), square it, and add it to both sides of the equation: (x^2  2x + (2/2)^2 = \frac{1}{3} + (2/2)^2) (x^2  2x + 1 = \frac{1}{3} + 1)

Simplify both sides of the equation: (x^2  2x + 1 = \frac{1}{3} + 1) (x^2  2x + 1 = \frac{4}{3})

Rewrite the left side of the equation as a perfect square: ((x  1)^2 = \frac{4}{3})

Take the square root of both sides of the equation: (x  1 = \pm \sqrt{\frac{4}{3}})

Solve for (x): (x = 1 \pm \sqrt{\frac{4}{3}})
So the solutions to the equation (3x^2  6x  1 = 0) using the completing the square method are (x = 1 + \sqrt{\frac{4}{3}}) and (x = 1  \sqrt{\frac{4}{3}}).
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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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