How do you solve using completing the square method #x^2+5x2=0#?
The square identity difference can be expressed as follows:
Color(white)(0) = (2x)^2 + 2(2x)(5) + 5^233#
((2x+5)sqrt(33))((2x+5)+sqrt(33))#color(white)(0)
Thus:
Thus:
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To solve the equation x^2 + 5x  2 = 0 using the completing the square method:

Move the constant term to the other side of the equation: x^2 + 5x = 2

To complete the square, take half of the coefficient of x (which is 5/2) and square it: (5/2)^2 = 25/4

Add this value to both sides of the equation: x^2 + 5x + 25/4 = 2 + 25/4

Rewrite the left side as a perfect square trinomial: (x + 5/2)^2 = 33/4

Take the square root of both sides: x + 5/2 = ±√(33/4)

Solve for x: x = 5/2 ± √(33/4)
Thus, the solutions are x = (5 ± √33)/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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