How do you solve using the completing the square method #x^2+4x+1=0#?
The solutions are:
To write the Left Hand Side as a Perfect Square, we add 4 to both sides:
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To solve (x^2 + 4x + 1 = 0) using the completing the square method, follow these steps:
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Move the constant term to the other side of the equation: [x^2 + 4x = -1]
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To complete the square, take half of the coefficient of (x) (which is 4) and square it: [\left(\frac{4}{2}\right)^2 = 4]
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Add this squared value to both sides of the equation: [x^2 + 4x + 4 = -1 + 4]
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Rewrite the left side as a perfect square: [(x + 2)^2 = 3]
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Take the square root of both sides: [x + 2 = \pm \sqrt{3}]
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Solve for (x): [x = -2 \pm \sqrt{3}]
So the solutions are (x = -2 + \sqrt{3}) and (x = -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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