Solve by Completing the Square x^2-2/3x=8/9

$x^{2} - \frac{2}{3} x = \frac{8}{9}$

Step 1

Simplify each term.

Combine $x$ and $\frac{2}{3}$ .

$x^{2} - \frac{x \cdot 2}{3} = \frac{8}{9}$

Move $2$ to the left of $x$ .

$x^{2} - \frac{2 x}{3} = \frac{8}{9}$

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 $b$ .

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

Step 3

Add the term to each side of the equation.

$x^{2} - \frac{2 x}{3} + {(- \frac{1}{3})}^{2} = \frac{8}{9} + {(- \frac{1}{3})}^{2}$

Step 4

Simplify the equation.

Simplify the left side.

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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

$x^{2} - \frac{2 x}{3} + {(- 1)}^{2} {(\frac{1}{3})}^{2} = \frac{8}{9} + {(- \frac{1}{3})}^{2}$

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

$x^{2} - \frac{2 x}{3} + {(- 1)}^{2} \frac{1^{2}}{3^{2}} = \frac{8}{9} + {(- \frac{1}{3})}^{2}$

Raise $- 1$ to the power of $2$ .

$x^{2} - \frac{2 x}{3} + 1 \frac{1^{2}}{3^{2}} = \frac{8}{9} + {(- \frac{1}{3})}^{2}$

Multiply $\frac{1^{2}}{3^{2}}$ by $1$ .

$x^{2} - \frac{2 x}{3} + \frac{1^{2}}{3^{2}} = \frac{8}{9} + {(- \frac{1}{3})}^{2}$

One to any power is one.

$x^{2} - \frac{2 x}{3} + \frac{1}{3^{2}} = \frac{8}{9} + {(- \frac{1}{3})}^{2}$

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

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{8}{9} + {(- \frac{1}{3})}^{2}$

Simplify the right side.

Simplify $\frac{8}{9} + {(- \frac{1}{3})}^{2}$ .

Simplify each term.

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{8}{9} + {(- 1)}^{2} {(\frac{1}{3})}^{2}$

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

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{8}{9} + {(- 1)}^{2} \frac{1^{2}}{3^{2}}$

Raise $- 1$ to the power of $2$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{8}{9} + 1 \frac{1^{2}}{3^{2}}$

Multiply $\frac{1^{2}}{3^{2}}$ by $1$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{8}{9} + \frac{1^{2}}{3^{2}}$

One to any power is one.

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{8}{9} + \frac{1}{3^{2}}$

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

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{8}{9} + \frac{1}{9}$

Combine fractions.

Combine the numerators over the common denominator.

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{8 + 1}{9}$

Simplify the expression.

Add $8$ and $1$ .

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

Divide $9$ by $9$ .

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

Step 5

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

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

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.

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

Any root of $1$ is $1$ .

$x - \frac{1}{3} = \pm 1$

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x - \frac{1}{3} = 1$

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

Add $\frac{1}{3}$ to both sides of the equation.

$x = 1 + \frac{1}{3}$

Write $1$ as a fraction with a common denominator.

$x = \frac{3}{3} + \frac{1}{3}$

Combine the numerators over the common denominator.

$x = \frac{3 + 1}{3}$

Add $3$ and $1$ .

$x = \frac{4}{3}$

Next, use the negative value of the $\pm$ to find the second solution.

$x - \frac{1}{3} = - 1$

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

Add $\frac{1}{3}$ to both sides of the equation.

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

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

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

Combine $- 1$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

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

Simplify the numerator.

Multiply $- 1$ by $3$ .

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

Add $- 3$ and $1$ .

$x = \frac{- 2}{3}$

Move the negative in front of the fraction.

$x = - \frac{2}{3}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = \frac{4}{3}, - \frac{2}{3}$

Solve by Completing the Square x^2-2/3x=8/9

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