Solve by Completing the Square 3x^2+2x+1=0

$3 x^{2} + 2 x + 1 = 0$

Step 1

Subtract $1$ from both sides of the equation.

$3 x^{2} + 2 x = - 1$

Step 2

Divide each term in $3 x^{2} + 2 x = - 1$ by $3$ and simplify.

Divide each term in $3 x^{2} + 2 x = - 1$ by $3$ .

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

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

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

Divide $x^{2}$ by $1$ .

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

Simplify the right side.

Move the negative in front of the fraction.

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

Step 3

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

${(\frac{b}{2})}^{2} = {(\frac{1}{3})}^{2}$

Step 4

Add the term to each side of the equation.

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

Step 5

Simplify the equation.

Simplify the left side.

Simplify each term.

Apply the product rule to $\frac{1}{3}$ .

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

One to any power is one.

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

Raise $3$ to the power of $2$ .

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = - \frac{1}{3} + {(\frac{1}{3})}^{2}$

Simplify the right side.

Simplify $- \frac{1}{3} + {(\frac{1}{3})}^{2}$ .

Simplify each term.

Apply the product rule to $\frac{1}{3}$ .

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = - \frac{1}{3} + \frac{1^{2}}{3^{2}}$

One to any power is one.

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = - \frac{1}{3} + \frac{1}{3^{2}}$

Raise $3$ to the power of $2$ .

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = - \frac{1}{3} + \frac{1}{9}$

To write $- \frac{1}{3}$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = - \frac{1}{3} \cdot \frac{3}{3} + \frac{1}{9}$

Write each expression with a common denominator of $9$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{1}{3}$ by $\frac{3}{3}$ .

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = - \frac{3}{3 \cdot 3} + \frac{1}{9}$

Multiply $3$ by $3$ .

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = - \frac{3}{9} + \frac{1}{9}$

Combine the numerators over the common denominator.

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = \frac{- 3 + 1}{9}$

Add $- 3$ and $1$ .

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = \frac{- 2}{9}$

Move the negative in front of the fraction.

$x^{2} + \frac{2 x}{3} + \frac{1}{9} = - \frac{2}{9}$

Step 6

Factor the perfect trinomial square into ${(x + \frac{1}{3})}^{2}$ .

${(x + \frac{1}{3})}^{2} = - \frac{2}{9}$

Step 7

Solve the equation for $x$ .

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

$x + \frac{1}{3} = \pm \sqrt{- \frac{2}{9}}$

Simplify $\pm \sqrt{- \frac{2}{9}}$ .

Rewrite $- \frac{2}{9}$ as ${(\frac{i}{3})}^{2} \cdot 2$ .

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

$x + \frac{1}{3} = \pm \sqrt{i^{2} \frac{2}{9}}$

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

$x + \frac{1}{3} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 2}{9}}$

Factor the perfect power $3^{2}$ out of $9$ .

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

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

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

Rewrite $i^{2} {(\frac{1}{3})}^{2}$ as ${(\frac{i}{3})}^{2}$ .

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

Pull terms out from under the radical.

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

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

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

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

Subtract $\frac{1}{3}$ from both sides of the equation.

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

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

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

Solve by Completing the Square 3x^2+2x+1=0

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