How do you solve #4x^2  x = 0# by completing the square?
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To solve (4x^2  x = 0) by completing the square:
 Move the constant term to the other side of the equation:
[ 4x^2  x = 0 ] [ 4x^2 = x ]
 Divide both sides by the coefficient of (x^2):
[ x^2 = \frac{x}{4} ]
 To complete the square, we need to add and subtract ((\frac{1}{2} \cdot \text{coefficient of } x)^2) inside the parentheses:
[ x^2  \frac{1}{4}x = 0 ]
Add and subtract ((\frac{1}{2} \cdot (\frac{1}{4}))^2 = (\frac{1}{8})^2 = \frac{1}{64}):
[ x^2  \frac{1}{4}x + \frac{1}{64}  \frac{1}{64} = 0 ]
[ x^2  \frac{1}{4}x + \frac{1}{64} = \frac{1}{64} ]
 Now, factor the perfect square trinomial:
[ \left(x  \frac{1}{8}\right)^2 = \frac{1}{64} ]
 Take the square root of both sides:
[ x  \frac{1}{8} = \pm \sqrt{\frac{1}{64}} ] [ x  \frac{1}{8} = \pm \frac{1}{8} ]
 Solve for (x):
[ x  \frac{1}{8} = \frac{1}{8} \quad \text{or} \quad x  \frac{1}{8} = \frac{1}{8} ]
[ x = \frac{1}{4} \quad \text{or} \quad x = 0 ]
So, the solutions to the equation (4x^2  x = 0) by completing the square are (x = \frac{1}{4}) and (x = 0).
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To solve (4x^2  x = 0) by completing the square, follow these steps:

Move the constant term to the other side of the equation: [4x^2  x = 0] [4x^2 = x]

Divide both sides by the coefficient of (x^2) (which is 4): [x^2 = \frac{x}{4}]

To complete the square, take half of the coefficient of (x) (which is (\frac{1}{4})), square it, and add it to both sides of the equation: [x^2  \frac{1}{4}x + \left(\frac{1}{4} \right)^2 = \frac{x}{4} + \left(\frac{1}{4} \right)^2] [x^2  \frac{1}{4}x + \frac{1}{16} = \frac{x}{4} + \frac{1}{16}]

Rewrite the left side as a perfect square trinomial: [\left(x  \frac{1}{8}\right)^2 = \frac{x}{4} + \frac{1}{16}]

To isolate (x), multiply both sides by 4 to clear the fraction: [4\left(x  \frac{1}{8}\right)^2 = x + \frac{1}{4}] [4x^2  x + \frac{1}{4} = x + \frac{1}{4}]

Subtract (\frac{1}{4}) from both sides: [4x^2  x = 0] [4x^2  x  \frac{1}{4} = 0]

Factor the left side, if possible: [(2x  \frac{1}{2})(2x + 1) = 0]

Set each factor equal to zero and solve for (x): [2x  \frac{1}{2} = 0] or [2x + 1 = 0]

Solve each equation for (x): [2x = \frac{1}{2}] or [2x = 1]
[x = \frac{1}{4}] or [x = \frac{1}{2}]
So, the solutions to (4x^2  x = 0) by completing the square are (x = \frac{1}{4}) and (x = \frac{1}{2}).
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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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