# How do you solve #x^2 + x + 1=0# by completing the square?

In this manner:

Since a square cannot be negative, the equation is impossible and has no solutions.

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This equation can't be solved by completing the square. Since D = b^2 - 4ac = 1 - 4 = -3 < 0, there are complex roots.

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To solve the quadratic equation (x^2 + x + 1 = 0) by completing the square, first, subtract 1 from both sides to isolate the quadratic terms:

[x^2 + x = -1]

Next, add ((\frac{1}{2})^2 = \frac{1}{4}) to both sides to complete the square on the left side:

[x^2 + x + \frac{1}{4} = -1 + \frac{1}{4}]

This simplifies to:

[(x + \frac{1}{2})^2 = -\frac{3}{4}]

To solve for (x), take the square root of both sides:

[x + \frac{1}{2} = \pm \sqrt{-\frac{3}{4}}]

Since the square root of a negative number is imaginary, we express it as:

[x + \frac{1}{2} = \pm \frac{i\sqrt{3}}{2}]

Finally, isolate (x) by subtracting (\frac{1}{2}) from both sides:

[x = -\frac{1}{2} \pm \frac{i\sqrt{3}}{2}]

So the solutions to the equation (x^2 + x + 1 = 0) are (x = -\frac{1}{2} + \frac{i\sqrt{3}}{2}) and (x = -\frac{1}{2} - \frac{i\sqrt{3}}{2}).

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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.

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