Solve by Completing the Square x^2-2/3x=4/9

$x^{2} - \frac{2}{3} x = \frac{4}{9}$

Step 1

Simplify each term.

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

$x^{2} - \frac{x \cdot 2}{3} = \frac{4}{9}$

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

$x^{2} - \frac{2 x}{3} = \frac{4}{9}$

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.

$x^{2} - \frac{2 x}{3} + {(- \frac{1}{3})}^{2} = \frac{4}{9} + {(- \frac{1}{3})}^{2}$

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

$x^{2} - \frac{2 x}{3} + {(- 1)}^{2} {(\frac{1}{3})}^{2} = \frac{4}{9} + {(- \frac{1}{3})}^{2}$

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

$x^{2} - \frac{2 x}{3} + {(- 1)}^{2} \frac{1^{2}}{3^{2}} = \frac{4}{9} + {(- \frac{1}{3})}^{2}$

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

$x^{2} - \frac{2 x}{3} + 1 \frac{1^{2}}{3^{2}} = \frac{4}{9} + {(- \frac{1}{3})}^{2}$

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

$x^{2} - \frac{2 x}{3} + \frac{1^{2}}{3^{2}} = \frac{4}{9} + {(- \frac{1}{3})}^{2}$

One to any power is one.

$x^{2} - \frac{2 x}{3} + \frac{1}{3^{2}} = \frac{4}{9} + {(- \frac{1}{3})}^{2}$

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

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{4}{9} + {(- \frac{1}{3})}^{2}$

Simplify the right side.

Simplify $\frac{4}{9} + {(- \frac{1}{3})}^{2}$ .

Simplify each term.

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

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

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{4}{9} + {(- 1)}^{2} {(\frac{1}{3})}^{2}$

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

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{4}{9} + {(- 1)}^{2} \frac{1^{2}}{3^{2}}$

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

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{4}{9} + 1 \frac{1^{2}}{3^{2}}$

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

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{4}{9} + \frac{1^{2}}{3^{2}}$

One to any power is one.

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{4}{9} + \frac{1}{3^{2}}$

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

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{4}{9} + \frac{1}{9}$

Combine fractions.

Combine the numerators over the common denominator.

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{4 + 1}{9}$

Add $4$ and $1$ .

$x^{2} - \frac{2 x}{3} + \frac{1}{9} = \frac{5}{9}$

Step 5

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

${(x - \frac{1}{3})}^{2} = \frac{5}{9}$

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.

$x - \frac{1}{3} = \pm \sqrt{\frac{5}{9}}$

Simplify $\pm \sqrt{\frac{5}{9}}$ .

Rewrite $\sqrt{\frac{5}{9}}$ as $\frac{\sqrt{5}}{\sqrt{9}}$ .

$x - \frac{1}{3} = \pm \frac{\sqrt{5}}{\sqrt{9}}$

Simplify the denominator.

Rewrite $9$ as $3^{2}$ .

$x - \frac{1}{3} = \pm \frac{\sqrt{5}}{\sqrt{3^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

$x - \frac{1}{3} = \pm \frac{\sqrt{5}}{3}$

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

$x = \pm \frac{\sqrt{5}}{3} + \frac{1}{3}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$x = \pm \frac{\sqrt{5}}{3} + \frac{1}{3}$

Decimal Form:

$x = 1.07868932 \dots, - 0.41202265 \dots$

Solve by Completing the Square x^2-2/3x=4/9

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