Solve by Completing the Square x^2+x+1=0

$x^{2} + x + 1 = 0$

Step 1

Subtract $1$ from both sides of the equation.

$x^{2} + x = - 1$

Step 2

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(\frac{1}{2})}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + x + {(\frac{1}{2})}^{2} = - 1 + {(\frac{1}{2})}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{1}{2}$ .

$x^{2} + x + \frac{1^{2}}{2^{2}} = - 1 + {(\frac{1}{2})}^{2}$

One to any power is one.

$x^{2} + x + \frac{1}{2^{2}} = - 1 + {(\frac{1}{2})}^{2}$

Raise $2$ to the power of $2$ .

$x^{2} + x + \frac{1}{4} = - 1 + {(\frac{1}{2})}^{2}$

Simplify the right side.

Simplify $- 1 + {(\frac{1}{2})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{1}{2}$ .

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

One to any power is one.

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

Raise $2$ to the power of $2$ .

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

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$x^{2} + x + \frac{1}{4} = - 1 \cdot \frac{4}{4} + \frac{1}{4}$

Combine $- 1$ and $\frac{4}{4}$ .

$x^{2} + x + \frac{1}{4} = \frac{- 1 \cdot 4}{4} + \frac{1}{4}$

Combine the numerators over the common denominator.

$x^{2} + x + \frac{1}{4} = \frac{- 1 \cdot 4 + 1}{4}$

Simplify the numerator.

Multiply $- 1$ by $4$ .

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

Add $- 4$ and $1$ .

$x^{2} + x + \frac{1}{4} = \frac{- 3}{4}$

Move the negative in front of the fraction.

$x^{2} + x + \frac{1}{4} = - \frac{3}{4}$

Step 5

Factor the perfect trinomial square into ${(x + \frac{1}{2})}^{2}$ .

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

Step 6

Solve the equation for $x$ .

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

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

Simplify $\pm \sqrt{- \frac{3}{4}}$ .

Rewrite $- \frac{3}{4}$ as ${(\frac{i}{2})}^{2} \cdot 3$ .

Rewrite $- 1$ as $i^{2}$ .

$x + \frac{1}{2} = \pm \sqrt{i^{2} \frac{3}{4}}$

Factor the perfect power $1^{2}$ out of $3$ .

$x + \frac{1}{2} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 3}{4}}$

Factor the perfect power $2^{2}$ out of $4$ .

$x + \frac{1}{2} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 3}{2^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} \cdot 3}{2^{2} \cdot 1}$ .

$x + \frac{1}{2} = \pm \sqrt{i^{2} ({(\frac{1}{2})}^{2} \cdot 3)}$

Rewrite $i^{2} {(\frac{1}{2})}^{2}$ as ${(\frac{i}{2})}^{2}$ .

$x + \frac{1}{2} = \pm \sqrt{{(\frac{i}{2})}^{2} \cdot 3}$

Pull terms out from under the radical.

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

Combine $\frac{i}{2}$ and $\sqrt{3}$ .

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

Move all terms not containing $x$ to the right side of the equation.

Subtract $\frac{1}{2}$ from both sides of the equation.

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

Reorder $\pm \frac{i \sqrt{3}}{2}$ and $- \frac{1}{2}$ .

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

Solve by Completing the Square x^2+x+1=0

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