How do you solve #3x^26x1=0# by completing the square?
Solution:
Either 3(x^22x+1) = 1 + 3 or 3(x1)^2 = 4#
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To solve the quadratic equation ( 3x^2  6x  1 = 0 ) by completing the square, follow these steps:

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

Divide both sides by the coefficient of the squared term to make the coefficient of ( x^2 ) equal to 1: ( x^2  2x = \frac{1}{3} )

Add and subtract the square of half the coefficient of the linear term (( x )) to complete the square: ( x^2  2x + ( 2 / 2 )^2 = \frac{1}{3} + ( 2 / 2 )^2 )

Simplify: ( x^2  2x + 1 = \frac{1}{3} + 1 ) ( x^2  2x + 1 = \frac{1}{3} + \frac{3}{3} ) ( x^2  2x + 1 = \frac{4}{3} )

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

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

Add 1 to both sides: ( x = 1 \pm \sqrt{\frac{4}{3}} )
So, the solutions 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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